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^ %. 



aV * 







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I 
I 
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if* < 




















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V 



■ 



THE ELEMENTS 



PLANE TRIGONOMETRY, 



WITH THE 



CONSTRUCTION AND USE 



LOGARITHMIC TABLES OF NUMBERS, 



AND THOSE OF 



TRIGONOMETRIC FUNCTIONS OF ANGLES. 



By J. C. SNOWBALL, M. A. 

FELLOW OF St. JOHN'S COLLEGE, CAMBRIDGE. 



SECOND EDITION. 



CAMBRIDGE: 

PRINTED AT THE PITT PRESS, BY JOHN SMITH, 
PRINTER TO THE UNIVERSITY. 

FOR T. STEVENSON, CAMBRIDGE; 
AND WHITTAKEr & CO., LONDON. 



M.DCCC.XXX1V. 




The Student is recommended to omit on the first reading, 

Articles 23. 24. 50 — 58. 67. 68. 70. 71. 76. 79- 82. 84. 
Cor. to 93. Cor. to 94. 97. 98- 109 — 111. 113—120. 124—128. 

Appendix I. 10 — 25. 
II. 10—16. 
III. 2. 3. 6 — 9. 11 — 19. 



-1 



PLANE TRIGONOMETRY. 



INDEX. 



CHAPTER I. 

ON THE METHOD OF REPRESENTING LINES AND ANGLES, 
AND THE DIVISIONS OF ANGLES. 

ART. PAGE 

1 — 3. Lines may be represented in magnitude and direction 

by Algebraical quantities 1 

4 — 8. The representation of angles 3 

10. The relation between the degrees and the grades which the 

same angle contains 5 

] 1, 12. Defs. The Complement, and the Supplement of an angle. 6 



CHAPTER II. 

OF GONIOMETRICAL FUNCTIONS OF ONE ANGLE, AND SOME FORMULAE 
CONNECTING THEM WITH EACH OTHER. 

13. Definition of Plane Trigonometry 7 

14 — 21. Definitions of the Goniometrical functions; — their 

changes of sign and magnitude 7 

22. Sin^ = ±sin(4w.90°±^), or = ± sin {(4ra + 2).90°=F A}. 13 

23. Tan ^4 = ±tan(4?i . 90°i A), or = ± tan {(4?z + 2).Q0° ± ^}. 14 

24 lCosA = cos(4w.90°±^), or= -cos{(4w + 2).90°±^}.l 
lSecJ= sec(4rc.90°i A), or = — sec{(4ra + 2).90°±^}-( 

25. Cot (90° - A) = tan A; cosec(gO° — A)=secA 16 

26 — 29. Formulae connecting together certain Goniometrical func- 
tions of one angle 16 

30. If A be less than 45°, sin A is less than cos A 20 

31. To find the sines, cosines, and tangents of 45°, 30°, 60° 20 



IV INDEX. 



CHAPTER III. 

GONIOMETRICAL FUNCTIONS INVOLVING MORE THAN ONE ANGLE. 

ART. PAGE 

32, 33. To find the sines and cosines of A ± B in terms of the 

sines and cosines of A and B 22 

34. To find sin 2 A and cos 2 A in terms of sin A and cos A 25 

36-38. ^ 0S ^ + SinJ = ± ^ + Sin ^H 25 

ICos^-sm A— ± v/C 1 -sm2^).j 

m , * Tix tan ^ ± tan B 

v 7 1 =F tan A . tan 25 

40 — 42. To express sin 2 A, cos 2 A, tan 2 ^ in terms of A 28 

43 — 47. Formulae connecting the sines and cosines of A, B, 

A±B 30 

lSmnA + s'm(n— 2) A = 2sin(n — l)A.cosA.) 
ICosw^ +cos(w — 2) A — 2 cos(?i - \)A.cosA.\" 

49. To find the sines and cosines of 1 8°, and 54° 33 

50. To determine some of the values of A which satisfy the 

equation, sin 2 A = cos 3 A 34 

51 — 54. To find the increments of the sine, cosine, tangent, and 
secant of an angle, when the angle receives a small 
increment 35 

55. A sin A = — A cos A, as cos A = sin A 36 

< < 

51. Tan- 1 /, ± tan~% = tan" 1 * l ** 38 

1 1 t j tg 

m , , *i — *2 1 *a — '3 

59. Tan- 1 ^ — tan -1 4= tan" 1 - — — - + tan -1 + .. . 

1 + t ! C 2 I + 1 2 f 3 

. . . + tan~ l - — , 39 

1 + t n -x • 4 



CHAPTER IV. 

ON THE SOLUT[ON OF TRIANGLES. 

.59 — fjS. Formulae used in the solution of triangles 40 

69— 71 . The solution of right-angled triangles 48 

72 — 79. The solution of oblique-angled triangles; and examples.. 51 



INDEX. V 

ART. PAGE 

80. To find the radii of the circles described in and about a 

regular polygon 60 

81. The area of a triangle in terms of the sides 6l 

. T „ sin B . sin C ,. 

82. Area of a triangle = \ a 2 . — = — -^ — ~ N 52 

& 2 sin (B + C) 

83. To find the radii of the circles described in and about a given 

triangle 62 

84. To find the area of a quadrilateral figure whose opposite 

angles are supplements to each other 63 

CHAPTER V. 

85 ON THE USE OP SUBSIDIARY ANGLES 65 



CHAPTER VI. 

ANALYTICAL TRIGONOMETRY. 

87- The circumference of a circle varies as its radius ............ 68 

(1YC 

88. — -p— is a proper measure of the magnitude of an angle 69 



(CLYC \ 
— -p — 1 of an angle to 

determine the number of degrees it contains ; and con- 
versely 70 

92. If x + - = 2 cos 6, then x = 2 V — 1 sin 6 72 

93. Demoivre's Theorem 72 

94. If 2 cos 6 = x + - , then 2 cos m 6 = x m +■ — , 

and 2 v/- 1 sin m 6 = x m 75 

95. 2 n - 1 .(cos^) n =cos?i0 + w.cos(w-2)O + «.^^-i.cos(w-4)a+... 76 

96. To express any positive integral power of the sine of an 

angle in terms of the sines and cosines of the multiples of 
the angle 77 

97. To expand cos nd in terms of cos0 ; n being a positive integer. 79 

98. To express cos/i0 in a series ascending by powers of cos0; 

n being a positive integer . . . , 81 



VI INDEX. 

ART. PAGE 

99- Having given tan 0, to find tan n 83 

100. If become 0, then — E - = 1, and —^- = 1 85 

U V 

101. To expand cos a and sin a in terms of a 86 

102. 103. The sine of a very small angle is equal to the angle 

itself very nearly 87 

~Cos0 = i.(eeV=l + e -eV=!) 
{ Sin d=±.V^i (eev=i - 6 -ev=i) 

1 e .ev-_ 1 

I an = - . . — 7 — = 

t/ - 1 e '^V-i + 1 

105. Gregorie's series; 0= tan — ^ — L + ^ an ' ... 89 

3 3 

106. Euler's series to determine 7r 91 

107- Machin's series; tt= 3*14159 ,... 91 

108. The tangent of a very small angle is equal to the angle 

itself very nearly 92 

109. Having given tan a, tan ft, tan y, ... to find tan (a + ft + . . . ) 93 

110. If ship = sin P. sin (z + p), to expand p in terms of sin P 

and the sines of z and its multiples 95 

111. If tan V = n . tan I, to expand Z' in terms of / 96 

112. To find the number of seconds contained in an angle which 

is expressed by the circular measure 97 

113. To expand . in a series of the cosines of and its 

r 1 — e.cosfl 

multiples 98 

114. To expand 4(1- e.cosfl) 100 

115—117. To expand (« 2 — 2ab.cos0+b*) n 101 

118. If sin (a> — y) = sin <o . cos n, to find the value of^y, which is 

small 1 05 

119- To find the approximate value of the small angle y from the 

equation cos {z + y) = sin n . sin z . cos m + cos z . cos n .* 1 05 

120. To expand, independently of Demoivre's principle, sinfl and 

cosO in terms of 1 08 



INDEX. 



Vll 



CHAPTER VII. 

ON THE SOLUTION OF EQUATIONS AND THE RESOLUTION OF 
CEBTAIN EXPRESSIONS INTO FACTORS. 

ART. PAGK 

121. To solve a quadratic equation , 110 

122, 123. To solve a cubic equation 113 

124—127. To resolve x 2n — 1, x 2n +l, x* n+1 _ 1, a? a+1 + l, into 

quadratic factors 114 

128. To resolve sin d and cos 6 into factors 121 



APPENDIX I. 

ON THE LOGARITHMS OF NUMBERS, AND THE CONSTITUTION 
AND USE OF LOGARITHMIC TABLES OF NUMBERS. 



1, 2. Definitions 123 

3. Having given tables of logarithms to any base to form tables 

to any other base 123 

4 — 6. Rules for performing multiplication, division, involution, 

and evolution by logarithms 124 

8 — 10. Description of the common, or Briggs', logarithms; — 

the advantages of having 10 for a base 125 

11, 12. Explanation of the arrangement of the common tables... 128 

13. Examples 129 

14. To expand l a (l + x) in a series ascending by powers of x. . . 130 
16 — 18. Series for the calculation of logarithms 133 

19. To expand a x in a series ascending by powers of x 135 

20. To find the base of the Napierian system of logarithms 137 

21. On the construction of the Tables of Logarithms 138 

22 — 25. On the construction and use of the small tables of pro- 
portional parts 138 

Examples 1 44 

26. The adaptation of formulae to logarithmic computation 145 

Two pages of Logarithms of Numbers 146, 147 



vm INDEX. 



APPENDIX II. 



ON THE CONSTRUCTION AND USE OF TABLES OF GONIOMETRIC 

FUNCTIONS. 

ART * PAGE 

1 — 8. The construction of the Tables 148 

9. Formulae of Verification 152 

1 1 . The increments of the tabular functions (mostly) vary as the 

increments of the angle 154 

12, 13. Use of the columns of N.D for l" 156 

14. The principle of Art. 11. not always practically applicable. . 158 

1 5. On the selection of a formula 159 

16. An angle less that 45° can be more accurately determined 

when its sine is given than when its cosine is given 159 

Page of certain Natural and Logarithmic Functions of Angles l6l 



APPENDIX III. 

ON THE CONSTRUCTION AND USE OF THE LOGARITHMIC 
TABLES OF GONIOMETRIC FUNCTIONS. 

1 — 5. On the determination from the angles of the logarithms 

of goniometric functions 162 

6. The increment of a tabular logarithmic function of an angle 

varies (mostly) as the increment of the angle 1 64 

7- To explain the meaning and use of the columns of differ- 
ences for 1" 165 

8. Examples 1 66 

9. The columns of differences for \" serve for increments for 

one set of functions and decrements for another set 168 

10. Rule for supplying the Tens when the Tabular Logarithms 

are used 1 68 

12 — 20. To determine very small angles from their L sines and 

L tangents ; and conversely 171 



PLANE TRIGONOMETRY. 

ALTERATIONS AND CORRECTIONS. 



PAGE 

8 
19 

46 

59 
81 
85 
94 
100 

101 

132 

171 



LINE 

5 



11 



9 

3 1 
lasti 

5 
2 

4 

17 



ERROR, 
if 



CORRECTION. 

is 
Vl-(cos^) 2 
cos-4 

360° ±4- 



for 2nd value of tan A 

360° + 4 

2 * 

in Prob. 5, the spectator is supposed to hold the string, 
after " integral powers" insert " of x". 

n-3 m-1 

(-1)— 



nearer to 
(l_vTT^)2 

(l + Vr^) 2 + e 2 
becomes e 




(l+Vl-e 2 ) 2 + e 2 

become e. 

dele "not". 



SPHERICAL TRIGONOMETRY. 

ALTERATIONS AND CORRECTIONS. 



>AGE 


LINE ERROR. CORRECTION. 


5 


5 from bottom after " AECO' 1 ' 1 insert " the angles of ". 




"equal to the" "angleof the". 


8 


1 Art. 10 Art. 9. 


40 


2 Art. 14. Cor. Art. 13. Cor. 



Since these pages have been printed, 1 have seen a neat and clear method of 

. . t • , t it • sm « sin ^4 

determining the tests which shew when the equation - 7 — 7 =- — — can be used to 

sm6 smi? 

determine any one of the quantities a, 6, A, B from the other three. It was 

my intention to insert with the list of Corrigenda, a geometrical proof shorter than 

that in page 40, but the analytical proof to which I allude, from which the 

following is taken, is much superior in conciseness, and quite as strict. I do not 

know who is its author. 

Instead of pages 40 to 43. 
Having given A, B, b, to determine in what cases a can be found from the 

sin a sin A 
equation -r~r = — — „. 
1 sin b smi? 



2 ' 



I. Let A + B be >tt. 
Then since, Art. 34, Cor., a+b and A + B are of like affection, 
/. a + b is >7r, 
or a>TT — b. 

If .*. b be <2, a ,is >-^. In this case, therefore, a may be found from 

- — r=- — ?r, and the triangle may be determined. 

sm6 sin£' & J 

If 6 be >^, we have nothing to enable us to determine from the relation 
a>ir — b whether a is > or <ir. 

II. If A + B < Tr, then a is < tt — b. 

And a is therefore < ij if b be > - , but cannot be determined if b be < - . 

III. If ^ + 5 = tt; then, Art. 34, (ix.), a + b = ir, and .'. a-ic-b. 
It appears then that A, B, b being given, 

when A + Bt>tt, and 6 < - ; a may be found, and it is > — , 

is i- 

u4 + 5<tt, and b>-^; a <"o, 

^4+5 = 7r; a = ir — b. 
And in no other case can a be found, and the triangle determined. 

Having given a, b, B, to determine when a can be found. 

If a + 6 = 7r, then as before A + B = ir, and .\ A = tt — B. 

< < < 

Whence we collect, as in the last proposition, 

when a + b>TT, and B < ~ ; A may be found, and it is > -^ , 

a + b<7r, and B>-^; A <2> 

a + b = Tr; A-tt — B. 
And in no other case can A be determined. 



St John's College, Sept. 12, 1835. 



PLANE TRIGONOMETRY 



CHAPTER I. 



ON THE METHOD OF REPRESENTING LINES AND ANGLES, 
THE DIVISIONS OF ANGLES. 



AND 



1. The magnitudes of lines may be represented by 
algebraical quantities. 

To measure a line we find how many times it contains a 
fixed and definite line, as an inch or a yard, which has been 
previously fixed on as the unit of length. Thus, if an inch 
be taken as the unit of length, we say that a line is SO, if 
it contain an inch thirty times : and in like manner we call 
a line a, if it contain the unit of length a times. 

2. If lines drawn in a given direction from a jioced line 
be represented by positive quantities, then those which are 
drawn from that line in a contrary direction will be repre- 
sented by negative quantities. 

If to a line AB measured from 
A to the right of the fixed line 
Ay it be required to add a given 
line, it is evident that AB must 
be produced to a point C such 
that BC shall be equal to that 

given line, — and if from A B we have to take a given line, we 
must cut off from BA a part BC' equal to the given line, 
and AC will be the required remainder. 

Let AB be represented by a and BC by 6, and make 
BC=BC; then 

AC + CB = AB; 
.-. AC = AB-CB = a -6. 
A 



c 



Now if b be greater than a, C lies on the other side of A. 
And AC = a — b = — (& — «), a negative quantity equal in 
magnitude to the difference of 
the lines b and a; which dif- 
ference we see lies in this case 
to the left of Ay. 

Hence, if a line be represented by a negative quantity, 
it is meant that it is taken in a direction from Ay contrary 
to that in which the lines represented by positive quantities 
were taken. 

3. Let XAX, YAY' be 

two fixed lines perpendicular to 
each other, and produced, if ne- 
cessary, to infinity. Then the 
position of any point P x in the 
plane of these lines is known 
if we know the magnitudes of 
the perpendiculars let fall from 
P l upon the lines XAX' and 
YAY', viz. P,N and P,M (or 
AN X , which is equal to P^M). 



:i 



Ml— l T> ' 



X' 



*s 



N a 



A. N A K__X 



Y'l 



A X N 13 PxNj, are called the co-ordinates of Pj referred 
to the rectangular axes AX and AY. 

With respect to lines measured from A in the direction 
of the axis XAX', it is usual to call those positive which 
are measured to the right of the axis YAY\ and therefore 
those will be negative which are measured to the left of YA Y' . 

With respect to lines measured in the direction of YAY 1 
it is usual to call those positive which lie on the upper side 
of XAX\ and therefore those will be negative which lie on 
the lower side of XAX' . 

Thus if P x be a point in the space contained by AX and 
AY, its co-ordinate AN X which is measured in the direction 
of XAX' is positive, because it lies to the right of YAY'; 
and its co-ordinate P Y N X measured in the direction of YAY' 
is positive, because it lies on the upper side of XAX'. But 




if P 2 be the point, its co-ordinate AN 2 measured in the 
direction of XAX' is negative, because it lies to the left of 
YAY' '; and the co-ordinate P 2 N 2 measured in the direction 
of YAY' is positive, because it lies on the upper side of XAX '. 

Similarly, the co-ordinates of P s and P 4 measured in the 
direction of XAX' are negative and positive respectively; 
and their co-ordinates measured in the direction of YAY' are 
negative in both cases. 

4. Def. If a straight line revolve in one plane round its 
extremity A from a fixed position as 
AB into any other position as AC, the 
inclination of AC to AB is called an 
angle (z); and the angle is signified 
by the letters BAC or CAB, the middle 
letter being that placed at the point 
in which the two lines meet. 

By continuing this revolving motion, we may suppose 
the angle to be made of any magnitude whatever. 

5. Def. If AD be equally inclined to the parts AB, 
AB' of the straight line BB', each of the angles BAD, BAD 
is called a right angle. 

6. Def. An acute angle is less, and an obtuse angle 
is greater, than a right angle. 

7- If the angles formed by AC revolving in one direc- 
tion, as BCD, from AB be considered positive, then if AC 
revolve in the contrary direction from AB it will form 
negative angles. 

If to the angle BAC, Fig. Art. 4, it be required to add a 
given angle, CA must move in the direction BCD through an 
angle CAE equal to the given angle, and the whole angle 
BAE will be that required. And if it be required to take 
a given angle from BAC, CA must evidently move in the 
contrary direction till it come into a position E'A, such that 
Z CAE' is equal to the angle to be subtracted. 

Then z CAE' + z BAE' = z BAC ; 
.-. lBAE' = z BAC - z CAE', 




Now if z CAE' be greater than 
z CAB, E'A lies on the other side of 
AB, 

and z BAE' = A BAC - Z CJ£' 
= -(zCXE'-z#JC), 

a negative quantity, whose magnitude is the difference be- 
tween the angles CAE' and BAC ; which difference we see 
lies in this case on the lower side of AB. 

Hence, by calling an angle negative, we mean that it is 
formed by the revolving line performing its revolution in a 
direction contrary to that in which it revolved to form posi- 
tive angles. 

8. A right angle is divided by the English into 90 equal 
angles called degrees ; a degree into 60 minutes ; a minute 
into 60 seconds ; a second into 60 thirds ; and so on to fourths, 
fifths, &c. And in the same manner as the length of a given 
line is determined by finding how many times it contains the 
unit of length, (as a yard), and its subdivisions, (viz. feet 
and inches) — so the magnitude of an angle is determined by 
finding how many times it contains a degree and its subdivi- 
sions. 

A degree and its subdivisions are thus marked, 24°, 50', 
34< ff , 42'" ; which denotes an angle containing 24 degrees, 50 
minutes, 34 seconds, and 42 thirds. 

In practice the subdivisions of the angle beyond seconds 
are not used ; if great accuracy be required, the smaller sub- 
divisions are expressed in decimal parts of a second : thus 
since 

42'" = ^ of l" = ^ = .7", 
60 60 

the angle above might have been written 24°, 50', 34" 7< 



9. By the French and other Continental Mathematicians, 
a right angle is divided into 100 equal angles called grades, a 
grade into 100 minutes, a minute into 100 seconds, and so on : 
and the divisions are thus marked 26', 24\ 32 M , 47"'. 

l' 
Since r» — =.01' 

100 



1' 



1 = 



100 



10000 

1* 



= .0001' 



r = 



= .000001'. 



100 1000000 
The above angle might have been written 26'-243247- 

Whence it appears that if the French division be adopted, 
arithmetical operations can be performed on angles in the same 
manner as on any other decimal fractions ; an advantage which 
does not attend the English division of the angle. 

10. To find the relation between E and F the number 
of degrees and grades contained in the same angle BAC. 
(Art. 14. Fig. 1.) 

E l BAC F 



90 


right z 




100 ' 


.-. E- 


10 


F 


F 

'To 9 


and F 


9 


E 


E 

+ -— . 
9 



Ex. 1. To find how many degrees and minutes are con- 
tained in the angle 42', 34', 56". 

F= 42.3456 





F 

10 ~ 


4.23456 


= F- 


F 

" 10~ 


38.11104 
60 

6.6624 
60 
39-744 


.-. E 


= 38°, 


6', 39"' 74. 



Ex. 2. Find how many grades and minutes are contained 
in the angle 24°, 5l', 45". 

First reducing the minutes and seconds to the decimal 
parts of a degree, 

60)45". 
60)51'. 75 









8625; 




. E = 


24 


.8625 




E 

9 


= 2 


.7625 


. E 


E 

+ — = 
9 


27 


.6250 



and .*. F=27 g , G2\ 50". 

11. Def. The complement of an angle is its defect from 
a right angle. 

Thus 

90° - 24°, 32' = 65°, 28' is the complement of 24°, 32', 
90° - 110°, 15' = - (20°, 15') is the complement of 110°, 15'. 

12. Def. The supplement of an angle is its defect from 
two right angles. 

Thus 180° - 56°, 20' = 123°, 40' is the supplement of 5&, 20', 

180° - 186°, 12' = - (6°, 12') is the supplement of 186°, 12'. 



CHAPTER II 



OF GONIOMETRICAL FUNCTIONS OF ONE ANGLE; AND SOME FORMULAE 
CONNECTING THEM WITH EACH OTHER. 



13. Def. Plane Trigonometry, in its original mean- 
ing, implies the measuring of plane triangles; — in its extended 
signification, it treats of the formulas connecting the relations 
of angles with each other, and of the solution of plane recti- 
lineal figures. 

14. Let a straight line revolve from the fixed line AB 
round the point A, and let there be A in the angle BAC. 



A T 



D' 



^ BIN: 



_B K 



Id' 



N B 



From any point C in AC draw CN perpendicular to AB, 
and through A draw DAD' perpendicular to AB. 

Now, for the reasons in Arts. 2 and 3, in these figures 
the signs of CN are +, + , -, - respectively, and those of 
AN are -f, — , — , + respectively. 

15. Definitions. See the figures of the last Article. 

CN CN 

l. -— - is the sine of the z BAC, or sin z BAC = — — . 
AC AC 

AN . AN 

2 - ~777 is tne cosine of the z BAC, or cos z BAC - — - . 
AC AC 



CN CN 

3. is the tangent of the z BAC, or tan z BAC- -j-rr 

AN AN 

4. — — - is the secant of the z 5JC, or sec z 5JC = -r— 

AN AN 

5. 1 - cos z iL4C is the versed sine of the /.BAC, 

or versin z 5JC = 1 - cos z Z?JC. 

6. The tangent of the complement of the angle BAC if ^ s 

the cotangent of the z BAC, 

or cot z BAC = tan (90° - z 5JQ. 

7. The secant of the complement of the z BAC is the 

cosecant of the z BAC, 

or cosec z 2?JC = sec (90° - A BAC J) 

8. The versed sine of the supplement of the z BAC is the 

suversine of the z BAC, 

or suversin z ^JC = versin (180° - z 5^4 C). 



16. The cosine of z i?JC might have been defined to be 
the sine of the complement of z BAC. 

AN 
For cos z BAC = — - = sin z ilCN=sin (90°- z .B^C). 

CN 
Also, sin BAC = — - = cos z JCiV = cos (90° - z #JC), 

or the sine of an angle is equal to the cosine of its complement. 

In the following pages we shall for the sake of conveni- 
ence indicate an angle by a single letter as sin A, cos A, tan B. 

17. Sin A, cos A, tan A... are proper measures to deter- 
mine the magnitude of an angle, since so long as the angle 
A remains the same, they are invariable whatever he the 
magnitude 0/ AC. 



Let D be any other point in AC, DM perpendicular to AB. 



Then by definition, 
sin A = 

or sin A = 



CN 
AC' 
DM 
AD ' 




N M B 



But by similar triangles ■— = — - ; or sin A is the same 

-aC JLU 

wherever in the line AC the point C be situated; — similarly 
it may be shewn that cos A, tan A, sec .4.. .are invariable so 
long as A remains the same. 

Hence, if any of the quantities sin A, cos A, tan A, sec A... 
be given, the angle A may be determined. 

2 
Ex. 1. Required to determine the angle whose sine is - . 

o 

Take any line AB and describe upon it a semicircle. 
From AB cut off the part BE equal to one third of AB, 
(Euclid vi. 9.), and with center A and radius AE describe a 
circle cutting ACB in C; join AC and CB. 

Then the angle ACB being in a semi- 
circle is a right angle, and 

sin J5C = —— = ■— - = - ; 
AB AB 3 

Wherefore ABC is the angle required. 




Ex. 2. Required the angle whose tangent is - . 

5 

Let a be any line and take AN = 5a, (fig. 1. Art. 14.) 
Draw NC at right angles to AN, and let it = 4>a : join ^4C. 

CN 4>a 4 



Then tan CJiV 



Wherefore CJiV is the angle required. 

B 



10 



18. Def. A quantity or expression of calculation is 
called a function of a quantity, if its value depend in any 
manner on the value of that quantity. 

Since the magnitudes of the sine, the cosine, the tangent, 
&c. of an angle depend on the magnitude of the angle, they 
are functions of the angle ; and when hereafter we make use 
of the term, the goniometrical functions of an angle, we mean 
those functions which are defined in Art. 15. 

19. To express versin A, cot A, cosec A, in terms of the 
sides of the triangle ANC. (Fig. 1. Art. 20.) 

AN 



(i) 



00 



Versin A = 1 — cos A = 1 — 



AC 



Cot A = tan (90° - A) 
= tan ACN 



AN _ . , „ _ 
= — — by del. of the tangent. 



Cosec A = sec (90° - A) 
= sec ACN 



CA 



= by def. of the secant. 

CN J 

20. To trace the signs of sin A, cos A, tan A, sec A, 
A increases from 0° to 360°. 



N 



5T 



B BIN 



B B' 



3V B 



CN 



(l) Sin A = -— , and therefore has the same sign in any 

■ALs 

case as CN has. 

Hence, (Art. 14.) sin A is positive if A be between 0° and 
180°, (figs. 1, 2.); and it is negative if A be between 180° and 
360°. (Figs. 3, 4.) 



11 



AN 

(2) Cos A = , and therefore has the same sign as AN, 

JxKs 

Hence, (Art. 14.) cos A is positive if A be between 0° and 
90°, or between 270° and 360°, (figs. 1, 4.) ; and it is negative 
if A be between 90° and 270°. (Figs. 2, 3.) 

CN 

(3) Tan A = — - , and is therefore positive or negative 

according as CN and AN have the same or different signs. 

Hence, (Art. 14.), tan A is positive if A be between 0° and 
90°, or 180° and 270°, (figs. 1, 3.) ; and it is negative if A be 
between 90° and 180°, or 270° and 360°. (Figs. 2, 4.) 

AC 

(4) Sec A — —pz 7 > ana * therefore has the same sign as AN. 

Hence sec A is positive if A be between 0° and 90°, or 
between 270° and 360°; and it is negative if A be between 
90° and 270°. 

21. To trace the magnitude of the sine, cosine, tangent, 
and secant, as the angle increases from 0° to 360°. (Figs. 
Art. 20.) 

Since (Art. 17.) the values of the goniometrical functions 
are not affected by the magnitude of AC, we will suppose 
this line to remain of the same magnitude while Z A increases 
from 0° to 360°. 

Now (fig. 1.) as CA revolves from AB to AD, CN 
increases from to AC and is positive ; and AN decreases 
from AC to and is positive. 

As (fig. 2.) CA revolves from AD to AB', CN decreases 
from AC to and is positive ; and AN increases from to AC 
and is negative. 

As (fig. 3.) CA revolves from AB' to AD', CN increases 
from to AC and is negative ; and AN decreases from AC 
to and is negative. 

As (fig. 4.) CA revolves from AD' to AB, CN decreases 
from AC to and is negative ; and AN increases from to AC 
and is positive. 



12 



Hence we collect, that as 



A changes from 


0° to 90° 


90° to 


180° 


180° to 


270° 


270° to 360° 


o +AC 

to 

AC AC 


+AC ■ 
to 




~AC 




—^ to 




-AC o 

to 

AC AC 


Mf)- 


+AC o 
AC '" AC 





-AC 


AC '" 




AC 


o +AC 


AC 


AC" AC 


Tm ,@... 


o + AC 

+AC'" o 


+AC 





-AC 





-AC 




-AC o 
o '"+-4C 


M9- 


AC AC 

+AC" o 





AC 
-AC 


-AC" 







The changes in sign and magnitude of the goniometrical 
functions may therefore be thus exhibited, the signs of the 
functions in each right angle being written in a bracket; 



A being 1 
between] ° 

sin A ..... 
cos A 

tan A ..... 
sec A ..... 



0° and 90° 



Oand 1,(+) 
1... 0, (+) 
o. ..«,(+) 

1 ... oo ,(+) 



90° and 180° 



1 and 0, (+) 
...-!,(-) 
oo ....Oj(-) 
oo ...-!,(-) 



180° and 270° 



0and-l,(-) 
-1 o,(-) 

0....oo,(+) 
-l....oo,(-) 



270° and 360° 



-1 and 0,(-) 

1,(+) 

oo ....0,(-) 



Since cos A is never greater than unity, versin A 
(or 1 - cos A) is always positive ; and its greatest value is 
when A becomes 180° or cos A becomes — 1, in which case 
versin A becomes 2. 



13 



% 


D 


c, 


r M 








TV 


°i 


W 


*A 



22. //* A 6e em?/ angle, and n be or any integer, then 
sin A = ± sin (4rc . 90° ± A), or = ± sin { (4<n 4- 2) 90° =f A ] . 
Let tfJd = J° ; and 54 d = B'AC 2 = J9'4C 3 = BAC 4 ; 

7l C_/ J — ZlCg = JJl\j 3 = .aC/49 

Then it may easily be shewn that, 
d C 4 and C 2 C 3 are each perpendicular 
to BB', 

C 1 N=C 2 M=C 3 M = C i N, 

and AN = AM. 

If the angle A be increased by 4^.90°, or AC X make any 
number {n) of complete revolutions so as to return into the same 
position, then it is clear that the sines of A and of 4w.90° + A 

are the same, viz. ; 

ACi 

or sin A = sin (4^.90° + J).... (1), 

Again, sin A = — ^— = -|— - = sin BAC 2 = sin (2.90 - J), 

-oIOjl Jl(^>2 
and by writing 2.90 — A for ^4 in (l), we have 
sin(2.90°-J)=sin [4>n. 90° + (2.90°- A)} = sin {(4^ + 2)90°-^}; 
.\ sin A = sin (2.90 -A) = sin {(4rc + 2) 90° - A\ (2.) 

C 3 M 



. . . A C t N -C 3 M 



4C S 



= - sin (2-90° + 4). 



And by (l), sin (2.90 + 4) = sin {4>n . 90° + (2.90 + A)} 

= sin { (4>n + 2) 90° + 4 } ; 
.\ sin 1 = -sin(2.90°+ A) = -sin {(4^ + 2)90° + ^} (3.) 

A . . dJV -C 4 N C,N . , 

Again, sin A = — = -— - = - — = - sm(4 .90° - 4). 

And by (l), writing m for n and 4.90° — A for 4, we have 
sin (4.90 - A) = sin {4m . 90° + (4„90° - 4)} 
= sin {4 (m + 1) 90° -4} 
= sin (4>n . 90° - A), if m + 1 = n ; 

.-. sin 1= ~sin(4.90° - 4) = - sin (4w.90°- A). .....(4.) 



14 

Hence from (1) and (4), and from (2) and (3), we have 

sin A - ± sin (4w.90° ± A) (i). 

sin A = ± sin {(4>n + 2) 90° =p J} (ii). 

Cor. If w = 0, we have from (i), sin A = - sin (- A). 

XT . C^ -C 4 JV C 4 i\T . 

Nowsmi= ^ = ^ = ^ a - smz ^ 

and BAC^ being in magnitude equal to BAC X or J, and lying 
below Ji?, will be represented by — A (Art. 7-) ; 

.*. sin A = - sin (- A), 

the same result as that given by the formula (i). 

Again, if n = 0, we have from (ii), 

sin A = sin (180° - A) 

sin A = - sin (180° + 1). 
If w = 1, we have from (i), 

sin A = sin (36o° + .A) 
= - sin (360° - A). 

23. If n be or any integer, then 
tan A = ± tan (4rc.90° ± ^), or = ± tan {(4<n + 2) 90° ± A}. 

If JC X (fig. Art. 22.) make any number (n) of complete 
revolutions, that is if A be increased by 4w.90°, we have 

C N 
tan A =-7T7 = tan (4^.90° + A) (l). 

AN w 

Again, 

¥ • C X N C 2 M C 2 M 

tan A = — - - = — = —_ = - tan BAC 2 

AN - AM AM 

= - tan (2.90° -A). 
And by (l), tan (2.90° - A) = tan {4rc.90° + (2.90 - A)} 

= tan {(4w + 2)90°- A} ; 
.\ tan J = - tan (2.90° - A) = - tan {(4rc + 2) 90° - J}... (2). 



15 

Again, 

t C,N -C 3 M C 3 M f n ,, 

tan A = ~ = — ^7 = -—r = tan { 2 -9°° + ^}> 

and by (l), tan (2.90° + J) = tan {4rc.90° + (2.90° + A)} 

= tan {(4^ + 2)90° + A], 
.-. tan J = tan (2.90 + J) = tan {(4rc + 2) 90° + A\ (3). 

Again, 

tan A = -J— = — i- = - -i— = - tan (4.90 - J), 

and by (l), tan (4.90 - A) 

= tan {4m. 90° + (4.90° - A)} 
= tan {4(m + l)90°- A} 
= tan (4w.90 -^), ifm+l=n; 
.-. tan .4 = -tan(4.90°- .4)= - tan(4w.90°- A) (4). 

Hence from (l) and (4), and from (3) and (2), we have 

tan A= ±tan(4w.90°± A) (i). 

tan A = ± tan {(4rc + 2)90° ± A} ... (ii). 

Cor. If n = 0, we have from (i), tan A — — tan (- A) ; — 
which appears also thus, 
. . CrN -C t N C t N ■ •■■ ■ ^ 

If w = 0, we have from (ii), tan A = ± tan (180° ± .4). 
If n = 1, we have from (i), tan A = ± tan (360° ± J). 

24. Similarly it may be shewn that 

cos A = cos (4rc.90°± J) (i), 

cos J = - cos { (4 n + 2) 90° =*= ^ } (ii), 

and sec A = sec (4^.90° ± J) (iii), 

sec J = - sec { (4 n + 2) 90° ± A\ (iv). 



16 

Cor. If n = 1, we have from (i) and (iii), 
cos A = cos (360° ± A), 
sec A = sec (s60° ± A). 

If ^ = o, we have from (i) and (iii), 
cos A = cos (- A), 
sec J = sec (- -4), 
and from (ii) and (iv), 

cos A = -cos (180° ± A) 9 
sec A = -sec (180° ± A). 

25. To shew that cot (90°- A)=tan A, and cosec (90°- A) 
= sec. A. 

For by definition, cot A = tan (90°- A), 
and writing 90°-^ for A, we have 

cot (90°- A) = tan \90°-(90°-A)} = tan A. 

Again, by definition, cosec A - sec (90° -A), 
and writing 90° -A for A, we have 

cosec (90° - A) = sec {90 - (90°- ^)}= sec J. 

26. It will be found necessary to remember the following 
expressions. 

CN 

(1) Tan^ = —— = — - = . 

v 7 AN AN cos A 

1c 



(2) Sec ^ = — — = 



AN AN cos A 
AC 



17 



AN 

J AN ~AC cos A 
(s) cotJ = _ = ^ = __ 

AC 



(4) cot A = — — 



(5) cosec A - 



CN CN_ tan A 
AN 

Ac - _L I 1 

CN~ CN~ sin J 
3C 



(6) AC°=CN°~ + AN 2 ; 

(CNy (AN\ 2 

or 1 = (sin il) 2 + (cos A) 2 . 

(7) ^C 2 = AN 2 + CN 2 ; 



CiW 



ir 



••' Ur) - * + {an) 

or (sec A)' — l + (tan J) 2 . 

27. By means of the expressions proved in the last article 
we can shew, 

sin d 

(1) tan A = . . 

y/l - (sin J) 2 

sin ^/i 
For tan J = — — by Art. 26, (l), 
cos A 



sin A 



by (6). 



(2) tan A 



VI -(sin J) 2 

sin ^4 \/l — (cos J) s 



18 



sin A tan A 

(3) sin A = 7 . cos A = 

7 cos A 1 

cos A 
tan A tan A 






sec J ^/i + (tan Af 
(4) cos J = \/l - (sin ^)* 



= vV(-4V 

\cosec J/ 

\/ (cosec J) 2 - 1 
cosec A 

28. By means of the values of the goniometrical func- 
tions deduced in Art 26, we can express one function in 
terms of any other. 

Thus, required the value of cosec A in terms of versin A* 
Cosec A ■■ 




Vl - (1 - versin A) 2 

1 
\^2 versin A — (versin A) 2 



29. In the first of the following Tables those values of 
sin A, cos A, tan A, and sec A are inserted which are the most 
useful to be remembered ; in the second Table are placed some 
others of less frequent occurrence in mathematical investigations: 



19 









CM 






^r 






^ 


^ 




.2 

CO 




^5 o3 




H 


O 

O 


^ 




CO 


y 


rH 


**■• 


rH 


CO v ^--' 




a 


^— ' 


•4-> 
Q 


i— i 


> 


CO 




1 




° 4- 
y + 




CO 


+ 


y 




1 






J^ 


^v—' 






J^ 












> 


> 






> 
















/7N 
















~A^ 


^ 








/"~*»\ 




/«»*■»-. /^»-> 








i-H 










CM 




CM ,— | 










5^ 




^ 


^ 


.2 

'35 


^ 


9 i 

CO « 




>j 




1 


53 


CO 

r* 


03 


^C 


i 


O 


T 8 





crj 




y 

CO 


^ 


> 


•«-> 


"co 


i 

> 


O 


rH S — ' 




-i-j 




O 


CO 

»H 

a 
> 


1 

T— ( 














> 






,—^-N 


^*^> 








^ 






<N 


•"->> 








/•^ 


1 




co 


.2 

"53 

V "-' / -H 


^5 






CO 


■4-J 

Q 


o 
y 


1 

y 


y 

y 

CO 


O 
o 


1 


+ 


CO 




O 
O 


O 


+ 


CO 

O 


o 
y 




v-V-w- 


JX- 








^^ 


e 






V 


■> 








> 


^ 
















CM 














9 




^ 


^— /^-^ 








| 


~*~> 


•p-i 




(N 












w 


CO 




^—*» 


^-^ '" H 










y""S 


£-1 




^ 


^ 1 










^ 




.2 

*co 


1 $ 


2 ^ 

T 8 

CO 


^5 

y 

CO 




.2 

*co 


- 


o 
y 

+ 


rH 


y 1 

8 2 




\^^J 


v^~* ^v^/ 












J_i 




> 


> > 










> 
















V^y^ 














^ 



20 



30. If A be less than 45°, cos A is greater than sin A. 
(Fig. 1, Art. 20). 

Let NAC be less than 45°. . 

Then, since NAC + iVd = 90°, iVCJ is greater than 45°. 

And in every triangle the greater side is opposite to the 
greater angle, (Eucl. i. 19.); 

.-. AN>CN; 
AN CN 
'• AC > At' 
or cos A > sin A. 

Similarly it may be shewn that for angles between 45° and 90° 
the cosine is less than the sine. 



31. To find the sines , cosines, and tangents of 45°, 30°, 
and 60°. 

(1) (Fig. 1. Art. 20). Let NAC = 45°; 

.-. NCA = 90° - NAC = 45° ; 

• \ AN = CN. And sin 45° = cos 45°. 

Also AC 2 = AN' + CN 2 = 2 AN"; 

AN 1 



.-. sin 45° = 



cos 45° 



and tan 45° 



AC yi 

71' 



sin 45* 
cos 45° 



= 1, 



(2) Let ABC be an equilateral and equi- 
angular triangle; therefore each of its angles 
is 1 of two right angles, or 6o°. 

Let AD be perpendicular to BC, — it bi- 
sects the angle BAC and the line BC ; 




21 



.-. BD = DC = iBC=±AB; and BAD=DAC=30 n ; 



. n BD \AB . 
' " AB~ AB »' 

cos 30° = v/ji-Csi 11 30f\ = ^/(l - i) 

. sin 30° 1 



v/i 



tan 30 L 



cos 30° ^/s 



(3) sin 60° = cos (90°- 60°), (Art. 16.) = cos 30° = 



V's 



cos 60° = sin (90°- 60°) 
sin 60° 



sin 30 



,o _ 1 



tan 60' 



,o_ 



cos 60 c 



\/s. 



CHAPTER III. 



GONIOMETRICAL FORMULAE INVOLVING MORE THAN ONE ANGLE. 



32. Having given the sines and cosines of two angles ; 
to final the sines and cosines of the angles equal to their sum 
and to their difference. 



BAC, CAD are two angles containing 
A and B d respectively. 

From D any point in AD draw 
DB, DC perpendiculars on AB and 
AC ; and from C draw CE, CF perpen- 
diculars on AB and DB. 

Then FE is a rectangle ; FB = CE, 
and FC = BE. 

L CDF = 90° - DCF = FCJ = ^, since CF is parallel to AE. 

xt • /; m DB BF+FD CE FD 

Now, sin (A + 5) = -— - = - — r — = + — _ 

v } AD AD AD + AD 

ce ac^ fd dc 
~ ac'ad* dc'Jd 




Also, cos ( A + 5) = 



= sin A . cos 5 + cos A . sin B (l). 

JJS _ AE - BE _ AE FC 
~AD~ AD ~~AD~^4D 



AE AC FC CD 
AC' AD " CD AD 



cos A . cos B - sin A . sin B 



(2). 



23 



Again, let z BAC = A, and Z CAD = B 

Fjom D any point in AD draw Z)^ and 
DC perpendiculars on AB and AC, CE a 
perpendicular from C on jfJ?, Z>JP per- 
pendicular to CE. 

FB is a rectangle ; i^£ = DB, and 
.FT) = EB ; 

z DCF=90° - ACE = A. 




Then, sin {A - B) = 



DB CE-CF CE CF 



Also, cos (v4 — B) 



AD AD AD AD 

_CE j c CF^CD 
~ AC' AD CD' AD 

= sin A .cos B — cos A . sin B (3). 

^B AE + EB AE FD 



AD 



AD 



AD AD 



AE AC FD CD 

+ 



AC AD CD AD 

= cos A . cos B + sin J . sin i? (4). 

Ex. Having given the sines and cosines of 45° and 30°, 
required the sine and cosine of 75°, and of 15°. 



Sin 45° = cos 45° = 



sin 30° 



, cos 30" = 



o y/s 



sin 75°^= sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° 



Art. 31. 



cos 75° = cos (45° + 30°) = cos 45° cos 30° - sin 45° sin 30 c 

V^2 2 \/V2 



24 

sin 15° = sin (45° - 30 a ) = sin 45° cos 30° - cos 45° sin 30° 

= — V ^- *- - 

\/2 ~¥~ a/2 ' 2 

= — ^(%/5-i); 

2\/2 
cos 15° = cos (45° - 30°) = cos 45° cos 30° 4- sin 45° sin 30° 

y/%' 2 \A' 2 

^(a/3 + I). 



2\/2 

33. tf one of the formulae, as sin ( A + B) = sin A cos B -f 
cos A sin B, be given, the others may be deduced from it. 

For let B become (- Z?), then 

sin (J-i?) = sin {^ + (-i?)}=sin A cos (-Z?) + cos A sin (-B). 

But cos (- B) = cos B ; Art. 24. Cor. 

and sin (- B) = - sin B ; Art. 22. Cor. 

.*. sin (A — B) = sin A cos 5 - cos A sin 2?. 

Again, 
cos (A + B) = sin {90° - (^ + B)} Art. 16. 

= sin {(90"- A) + (-B)\ 

= sin (90° - A) cos (- 5) + cos (90° - A) sin (- B). 

But sin (90° - A) = cos J, and cos (90° - A) = sin ^4 ; Art. 16. 

.-. cos ( A + J5) = cos A cos J5 - sin A sin 5. 

Again, 
cos (A-B) = sin {90° - (J - B) \ 

= sin {(90°-^) +B} 

= sin (90° - A) cos J5 + cos (90° - A) sin 5 

= cos A cos /? + sin A sin #. 



25 

34. To shew that sin 2 A = 2 sin A cos A. 

Sin (A + B) = sin .A cos Z? + cos J sin B, 
and writing ^f for B this becomes 

sin 2 A = sin J cos A -f cos J sin J. = 2 sin J cos A* 

35. jTo s^ew that (l) cos 2 A = (cos A) 2 — (sin J) 2 . 

(2) cos 2 A = 2 (cos J) 2 - 1. 

(3) cos 2 Jl = 1 - 2 (sin J) 2 . 

(1) Cos ( A + 2?) = cos J cos Z? - sin A sin i?, 

and writing A for J5, 
cos 2 A = cos ^ cos J - sin J sin ^( = (cos J) 2 - (sin J) 2 « 

Again, 

cos 2 A = (cos J) 2 - (sin J) 2 } 

And 1 = (cos A) 2 + (sin A) 2 ; 

.*. 1 + cos 2 A - 2 (cos A) 2 , 

1 -cos 2 A = 2 (sin A) 2 . 

(2) Therefore cos 2A - 2 (cos J) 2 - 1. 

(3) And cos 2 A = 1 - 2 (sin 4) 2 . 

36. To s^ew #Aa£ cos A + sin A = ± y/(\ + sin 2 A), 

cos A - sin A = ± ^/(l - sin 2 A). 
Sin 2 A = 2 sin A cos A, 
And 1 = (cos A) 2 + (sin A) 2 , 
/.by addition and subtraction, 

1 + sin 2 A = (cos J) 2 + 2 sin J cos J + (sin .4) 2 , 
1 - sin 2 ^( = (cos J) 2 - 2 sin A cos J + (sin A) 2 ; 

.-. cos /4 + sin A = ± /^/(l + sin 2^), 

and cos A - sin J = ± v^C 1 "" sm 2 ^)* 
D 



26 

37- To shew that if A be less than 45°, 

(cos A = \ {\/(l + sin 2 A) +o/(l -sm2A)\, 
then -J 

I sin J = 1 { ^/(l + sin 2 J) - ^/(l - sin 2A)}. 

By Art. 30, if J < 45°, cos A is > sin J, and they are 
both positive ; 

.-. cos A + sin A = + ^/(l + sin 2 J), 

and cos A — sin ^4 = + >y/(l - sin 2 A) ; 

.\ by addition and subtraction, 

2 cos J. = /y/(l + sin 2 A) + ^/(l - sin 2 J), 

2 sin A = ^/(l + sin 2 A) - ^/(l - sin 2 A), 
.\ cos J = \ \\/(l + sin 2 A) + ^/(l - sin 2 A)}, 
sin J = 1 {-\/( l + sm % A) - \/(l -sin 2^)£. 

38. The equations deduced in Art. 36 can be applied in 
any case to determine the sine and cosine of A from sin 2 A. 

For example, if A be an angle > 3 x 45° but < 4 x 45°, 
(i. e. any angle comprehended under the form 180° - B, 
where B is < 45°), cos A is negative and greater in magni- 
tude than sin A which is positive. 

In this case therefore, 

cos A + sin A = - ^/(l + sin 2 A), 

cos A - sin A = - ^/(l - sin 2 J) ; 

.-. cos A =-\ {a/0 + sm 2 ^) + a/0 ~ sm 2 ^M' 
sin J = ^ \ ^/(l - sin 2 J) - ^/(l + sin 2^)}. 

If at first sight this value of sin A appear to be negative, 
it is to be remembered that sin 2 A is a negative quantity, 
(2 A being between 270° and 360°) and therefore the value of 
sin A is positive as it ought to be. 



27 



39- Having given the tangents of two angles to find the 
tangents of their sum and their difference. 



Tan (A + B) 



sin (J + B) 
cos {A + B) 

sin A cos B + cos A sin 5 
cos A cos 5 — sin A sin 5 



and dividing numerator and denominator by cos A cos B, 



sin J sin B 

+ 



cos ^ cos B 



] 



sin J sin B 



Similarly, tan (A - B) = 
Cor. 1. If B = 1, tan 2 J = 



cos ^4 cos 2? 

tan ^4 + tan i? 
1 - tan A tan 5 

tan A - tan B 



1 + tan A tan 5 
2 tan A 



1 - (tan J) 2 

Con. 2. If 5 = 45°, since tan 45° = 1, (Art. 3l) 

/ \i ^ 1 + tan^ /-x 

tan (J + 45°) 



Similarly, tan (^ - 45°) = 



1 -tan J " 






sin A 
cos A 






sin -4 
1 — 






cos A 




cos J + sin A 


....(2). 




cos J - sin A " 




tan J - 1 , x 


sin A - 
r — 


- cos A 



tan A + l 



sin A + cos A 



,(*>, 



/ a nx / a nv 1 + tan J tan A - l 

tan (A + 45°) + tan (J - 45°) = 4- z 

v ' ' 1 - tan A tan A + 1 

4 tan A 



1 - (tan Af 

= 2 tan 2 J, by Cor. 1.... (5). 

If A be<45°; 

Since tan (A - 45°) = tan - (45° - A) = - tan (45° - A), 

(Art. 23. Cor.) 
the last expression becomes 

tan (45° -M) - tan (45° - A) =2 tan 2 J, (6). 

40. To find the value of sin 2 A and cos 2 A in terms 
of tan A. 

Sin 2 J = 2 sin J cos A, Art. 34. 

2 sin A , 



cos A 


<"■> 


2 tan A 
~ (sec Af ' 


Art. 26. (2) 


2 tan J 
1 + (tan A] 


- , Art. 26. (7). 

> 2 


Again, cos 2 A = 2 (cos J) 2 - 1, Art. 35. (2) 


2 

— i 


(sec 


Af 




2 


l + 


(tan J) 2 


1 - 


(tan Af 



1 + (tan Af 

41. The following values of sin 2^4, cos 2 A, tan 2 -4, are 
of frequent occurrence, and necessary to be remembered ; those 
of them which have not already been proved are easily deduced 
after the method used in the last Article. 



29 



I. Sin 2A=2 sin A cos A. 

2 tan A 
2 



1 + (tan A)' 



4. Tan2l : 



2 A /{(secJ) 2 -l 
(sec A) 2 

2 tan i. 



1 - (tan Af 



I. Cos 2 1= (cos .4) 2 -(sin l) 2 

2 = 2 (cos A) 2 -1. 

3 =1-2 (sin J) 2 . 

1 - (tan A) 2 
4 " l+(tan.A) 2 ' 

2 - (sec .i) 2 
(sec A) 2 



42. In the same way can be found the following 
values of sin 2 A and cos 2 J in terms of cot A, cosec J, and 
versin A. 



Sin 2 1 



sin 2 A = 



2 cot .A 



l + (cot A)" 



2 a/{ (cosec Af -1} 



(cosec A) 2 

sin 2 J = 2 . {l - versing} \/{ 2 vers ^ ~ (vers^) J 
(cot A) 2 - 1 



Cos 2 A 



cos 2 A 



(cot l) 2 + 1 
(cosec J.) 2 — 2 



(cosec A) 2 

cos 2^4 = 1 -2 {2 vers 1 - (vers^4) 2 }. 

It may be remarked that the easiest method of deducing 

formulae such as these, is first to express sin 2 A and cos 2 A in 

terms of sin A and cos A . Thus let it be required to prove 

that 

(cosec A) 2 — 2 

cos2A = ~ '—-r- ; 

(cosec A) 

cos 2 A = (cos A) 2 - (sin A) 2 

= 1-2 (sin A) 2 ; 



30 



But cosec A = - — r, or sin A ; 

sin A cosec A 

2 (cosec A) 2 - 2 

.'. COS 2 A = 1 - — — = —3 jrr— 

(cosec .4 ^ (cosec A)^ 



43. Since sin (A + 5) = sin A cos i? + cos 1 sin B, 

and sin {A - B) - sin 1 cos 5 - cos A sin 5 ; 
.-. by adding and subtracting, we have 

sin (A + B) + sin (A - B) = 2 sin J cos Z? (l). 

sin (J. + 5) - sin (4 - #) = 2 cos A sin J? (2). 

Similarly it may be shewn that 

cos (A + B) + cos (A - B) = 2 cos A cos J? (3). 

cos (A - B) - cos ( J + 2?) = 2 sin A. sin i? (4). 

44. To find the values of sin A ± sin B and cos A ± cos B, 
in terms of the sines and cosines of ^ (A + B), and 1 (A - B). 

Since A = \ {A + 5) + 1(^ - B), 

and 5=l(J+5)-l(^-J5); 

.-. sin A = sm^(A+B)cos±{A-B) + sin±(A-B)cos±(A+B), 

sin 5 = sin \ (A+B)cosi>(A-B) - sin £ (A-B) cos±(A+B) ; 

.-. sin J +sinB = 2sin±(A + B) cos 1(1 - B) (l). 

sin^f - sini? = 2cosl(J + B) sin 1 (A - B) (2). 

Similarly, 

cos A + cos B = 2 cos 1 (A + B) cos 1 (J -5) (3). 

cos /? - cos A = 2 sin 1 (^ + #) sin 1 (i - B) (4). 



31 

These four formulae might have been deduced from the for- 
mulae of the last Article, by making 

A+B = S, and A - B = D, 

in which case 

A = ±{S + B), S = i(S-D). 

45. The formulae of the two last Articles are necessary 
to be remembered; the following are not of such frequent 
occurrence. 

Dividing (2) by (l), Art. 44. 

sin A - sin B _ 2 cos \ ( A + B) sin |( A - B) _ tan \ (A - B) 
sin A + sin B ~ 2 sin A (A + B) cos 1 {A - B) ~ tan ^ (A + B) 

Similarly, dividing (4) by (3), 

cos B — cos A 



cos J+ cos 2? 

So also 

sin A ± sin B 



cos A + cos B 



tan 1 (J + B) tan 1 (J - B). 
tan 1 (^ ± B) ; 



sin J. ± sin B 
co Sj g-cos^ =COtan ^ (jl:F ^ 

46. Ta„^ ± ta„ J B=^i ± si 



cos A cos i? 

sin A cos 2? ± cos A sin 5 
cos il cos B 

sin (^ ± B) 



cos .4 cos 5 

„. ., , „ „ sin (j4 + 5) 

Similarly, cot J + cot i? = : — - 

sin i.sm5 

sin (1 - 5) 

cot B - cot A - -r— : — - . 

sin y^ . sin B 



32 

4tf> Sin (A + B).sin(A - B) 

= (sin Af (cos Bf - (cos Af (sin Bf 

= (sin J) 2 {l - (sin B) 2 } - {l - (sin A) 2 } (sin fl)» 

= (sin J) 2 - (sin #) 2 . 

Similarly, sin (A + B) sin ( A - B) - (cos B) 2 - (cos J) 2 . 

In like manner it may be shewn that 

cos (A + B) cos ( A - B) = (cos J) 2 - (sin Bf 
or, =5 (cos B) 2 - (sin A) 2 . 

48. From (l) and (3) of Art. 43, (w - l) J being written 
for A 9 and J for i?, we have 

sinnA + sin (n - 2) A = 2 sin (w - 1) A . cos J (l). 

cosnA + cos (w - 2) J = 2 cos (^ - 1) A . cos A (2). 

If n = 2, from (l), 

sin 2 J = 2 sin J cos J., 
from (2), cos 2 A + 1 =2 (cos .4) 2 , 

or cos 2 J = 2 (cos ^) 2 - 1. 
If n = 5, from (l), 

sin 3A = 2 sin 2 J cos J - sin ^ 
= 4 sin J (cos J) 2 - sin A 

- 4 sin J {l - (sin A 2 )} - sin J 
= 3 sin J — 4 (sin Af. 

From (2), 

cos 3A = 2 cos 2 J cos A — cos A 

- 2 cos ^4 {2 (cos J) 2 - 1 } ~ cos A 

- 4 (cos J) 3 - 3 cos J! ; 

and by successive substitutions sin 4 A, sin 5 ^... cos 4<A, cos 5 A,.. 
might be found. 



33 

49. To find the sines and cosines of 18° and 54°. 
Sin 36° = cos (90° - 36°) = cos 54°, 
or, if 18° = A, sin 2 A = cos 3 A ; 

.-. 2 sin J cos J = 2 cos 2 J cos A - cos A, Art. 48. 

= 2 cos A \ 1 - 2 (sin J) 2 } - cos J ; 

.-. 2 sin J = 2 - 4 (sin J) 2 - 1 ; 

.-. 4 (sin J) 2 + 2 sin J = 1 ; 

And solving this equation we have, 

. ±\/5-l 
sin A — • 

4 

Where the positive sign is to be taken because sin 18° is a 
positive quantity; — the negative sign indicates other angles, 
(whose value will be determined in the next Article,) which 
satisfy the equation sin2^ = cos3^; 

.-. sin 18° = — — Z— = cos (90° - 18°) = cos 72°...(l). 

6-2\/~5 10 + 2\/5 

(cos 18 ) 2 = 1 - (sin 18 ) 2 = 1 — = — - — — , 

V J K J 16 16 

.-. cos 18° = — — I = sin 72° (2). 

4 

Again, sin 54° = cos 36° = cos 2.18° = (cos 18 ) 2 - (sin 18 ) 2 

_ 10 + 2<s/~5 6 -2o/~5 
= 16 16 

-^ ■■ » 

/ ,.0X2 r- n,9 6+2^/5 10-2\/5 

(cos 54°V = 1 - (sin 54 ) 2 = 1 — = — , 

' V J 16 16 

.'. cos54°= ^-A— Z—l = sm3 Qo ^y 

E 



34 

50. To find some of the values of A which satisfy the 
equation, sin 2 A = cos 3 A. 

cos 3 A = sin 2 A = cos (90° - 2 A) 

= cos {4n . 90° ± (90° -24)} Art. 24. (i). 

= cos {(4w± 1)90°=f2^. 

Let 3 A = (4w ± l) 90° =f 2 J ; 

.*. 3 J ± 2^ = (4w ± 1) 90°. 

Taking the upper sign, 

90° 
A = (4w + 1) = (4rc + 1) 18° (1). 

5 

Taking the lower sign, 

A = (4<n - 1)90° (2). 

Again, 
cos 3 A = sin 2 J = cos (90° - 2 J) 

= - cos \(4,n + 2) 90° ± (90° - 2 A)} Art. 24. (ii). 

= cos { 180° ± [(472 + 2)90° ±(90° -2 A)]} Art. 24. Cor. 

Let 3 A = 180° =fe |(4 w + 2) 90° ± (90° - 2 A)} ; 
.-. 3 ^f ± 2 J = 180°± (4w + 2) 90° ± 90°. 

And by solving this equation, we obtain other values of 
A which satisfy the proposed equation. 

For n writing 0, 1, 2, 3 successively, we have 

from(l); A= 18°, = 5.18°, = 9-18°,= 13.18°,... 

from (2); A= - 90°, = 3-90°, = 7-90°, = 11. 90°,... 



35 

Con. The angle 13.18°, or 234°, is one of those alluded 

\/5 + 1 

to in the last Article whose sines are - . 

4 

For sin 234° = sin (180° + 54°) 

= - sin 54° Art. 22. Cor. 

= - 1+V ^. Art. 49. (4). 

4 

51. If an angle receive any increment , to find the 
corresponding increment of the sine of the angle. 

Let A receive an increment $A, and let the correspond- 
ing increment of sin A be represented by AsinX 

Then A sin A 
= sin (A + $A) - sin A 
= (sin A cos 5 A + cos A sm%A) - sin A 
- cos A sin S J + sin A . (cos S A - l) 
= cos^sin£J+sinJ{[l-(sin£j) 2 ]i-l} Art. 26. (6). 

= cosJsin£j + sin,l{[l-l(sinSJ) 2 (sin£j) 4 -...]-l} 

tii . *± 

= cos^.sin$J-sin.^{l(sin<L4) 2 H (sin^i) 4 +...} 

= cos^sin$J{l-tan J[lsin<L4 + (sin £j) 3 +...]} 

= cosJsin^J{l-ltanJsin^J[l+i(sin^) 2 -|-...]}. 

Cor, 1. If SA be a very small angle, and therefore 
sincLi be very small, J (sinSA) 2 and all the terms which 
follow it, being higher powers of sin^il, may be neglected 
when compared with unity, and we have 

A sin J = cos A sin § A (l - \ tan A sin I A). 

Cor. 2. Also, $ A being very small, — unless tan A be very 
large, that is, unless A be nearly (2n+ 1) 90°, — 1 tan A sin §A 
may be neglected when compared with unity, and we have 

A sin A = cos A sin $A* 



36 

52. If an angle receive any increment, to find the cor- 
responding increment of the cosine of the angle. 

A cos A = cos {A + 8 A) — cos A 

= cos A cos 8 A - sin J sin 8 A - cos A 

= — sin A sin 5 A + cos ^4 (cos 8 A — 1) 

= - sin A sin(L4+cosJ{[l-(sincL4) 2 ]^-l} 

= - sin A sinc)J-cosJ{l(sincL4) 2 + (sincLf) 4 +...} 

= -sin^.sin^{l+lcotlsin^l[l+l(sin £J) 2 +...]}. 

This quantity, being negative, shews that as an angle 
increases its cosine decreases. 

Cor. 1. If 8 A be very small, we have, as in Cor. 1. of 
the last Article, 

A cos A = - sin A sin 8 A (l + -| cot JsincM). 

Cor. 2. Also, if 8 A be very small, and cot A be not very 
large,— that is, if A be not nearly 2 n . 90°, — we have, as in 
Cor. 2. of the last Article, 

A cos A = - sin A sin 8 A. 

53. If an angle receive any increment, to find the 
corresponding increment of the secant of the angle. 

A sec A 

= sec (A + 8 A) - sec A 
1 1 



cos (A + 8 A) cos A 

cos A - cos (A + 8 A) 

cos A cos (J + 8 A) 

- {cos (J + 8 A) - cos .A } 

cos A (cos ^ cos $A - sin it sin 8 A) 

sinlsin<!U{l+^cotlsin£j[l+^(sincL4) 2 +...]} At r 2 

(cos A) 2 cos 8 A (l - tan A tan £A) 

^l+Acotlsin^A [l + i(sincL4) 2 + ...] 

tan A sec A tan d J * L * . 

1 - UxnA tan dA 



37 

Con. 1. If SA be very small, 

. 1 - 1 cot A sin 5 J 

A see A = tan J sec A tan d 4 - -—. . 

1 - tan A . tan 6 A 

Cor. 2. If 5J being very small, neither tan A nor 
cot A be very large, — that is, if A be not n . 90° nearly, — - 
then both 1 cot A sin S A and tan A tan S A may be neglected 
when compared with unity, and we have 

A sec A = tan A sec A tan SA. 

54. //* an angle receive any increment, to find the cor- 
responding increment of the tangent of the angle. 

A tan A = tan (A + S A) - tan A 

sin ( A + S A) sin A 
cos (J + $ A) cos A 

sin (^4 + SA) cos J - cos (A + 5 J) sin A 
cos ^ (cos -4 cos SA — sin ^ sin SA) 

But sin (A + SA) cos J - cos (A + SA) sin A 

= sin {(-4 + 5-4) - A} - sinSA; 

.-. A tan J = 



sin 5 J 



(cos J) 2 . cos S A (l - tan J tan $ J) 
(sec A) 2 . tanSA 



1 - tan A tan 5 J 



Con. If 5-4 being very small, tan A be not very large, - 

that is, if A be not (2n + 1) 90° nearly, — we have 

A tan A = (sec A) 2 tan S A. 

55. .For a small increment of A, £Ae increment of the 

> 

sine is = the decrement of the cosine, according as cos A 



is = sin A ; A not being very small or nearly a multiple of 90°. 



38 

For Asin A = cos J sin 8 A, if A be not nearly (2w + 1)90 (} . 

Art. 51. Cor. 2. 

A cos A = - sin A sin 8 A, if A be not nearly 2ra . 90°. 

Art. 52. Cor. 2. 

Hence, if ^4 be not very small, or nearly a multiple of 90°, 

> > 

A sin A is = (- AcosJ), as cos J is = sin A. 



Cor. In angles less than 90°, A sin A is (- A cos A) 



< 



as i is 45°. Art. 30. 



56. Def. By tan l t is meant the angle whose tangent is t. 

So by sin _1 s and cos _1 c, &c. are meant respectively the 
angle whose sine is s, and that whose cosine is c, &c. 

57- To shew that tan" 1 ^ + tan" 1 4 = tan 



1 -Ms 
Let tan A = t x and tan B = t 2 . 

Then, by definition, A = tan~ 1 t l , and ^=tan' -1 ^. 

tan A + tan B 



Now tan (A + 5) 



1 - tan A tan 5 

tan A + tan Z? 



.'. by def. J + B = tan M 

J \\ - tan J tani? 

£i + t 2 
Or tan " 1 t x + tan ' 4 = tan 



1 -t x t % 



Similarly it may be shewn that 

tan -1 t x - tan~% = tan^ 1 



+ tj 2 



39 



58. If t 1? t 2 ,...t n be the tangents of any angles, 



_ i *i ' "~ H _ i H "~ *3 i £» - 1 — £ 

= tan - + tan —7 + . . . + tan 



1 + t y t c 



1 + Ut, 



1 +t n _ l t n 



For, tan 1 # 1 -tan ^ 2 = tan 



tan l L — tan *£, = tan 



_! ^1 ~ *2 

' 1 + Ms 

.1 H ~ H 
1 + £>£, ' 



tan 1 ^ M _,-tan T £„=tan~ 



1 + * n -l*n 



by addition, tan l t y — tan 1 t 7i 



= tan 



1 + *;£> 



+ tan 



*2 ~ ^3 
1 + t 2 t. 



- +... + tan 



1 + t n _it n 



The Appendices 1, 11, in, on the Logarithmic Tables 
of Numbers and those of Goniometrical Functions, ought to be 
read before entering on the next Chapter. 



CHAPTER IV. 

ON THE SOLUTION OF TRIANGLES. 



59. A triangle consists of six parts, namely, three 
sides and three angles ; when three of these parts are given 
(except they be the three angles) it will be shewn that the 
other three can, generally, be found. 

The number of degrees in the angles of a triangle will be 
designated by the letters A, B, C, which are placed at the 
angular points of the triangle, and the sides respectively oppo- 
site to the angles A, B, C by the letters a, b, c. 

60. The sines of the angles of a triangle are propor- 
tional to the sides respectively opposite. 

Let ABC be the triangle, and draw CD perpendicular 
to AB, or AB produced either way. 




CD 

Then sin CAB = sin CAD = — - . 

CA 

(With reference to the third figure, see Art. 22. Cor.) 

CD 



Also sin CBA = sin CBD = 



sin CAB CB 



CB' 



Or, 



sin CBA CA 
sin A a 



smB b 

And A, B are any two angles of the triangle. 



41 

sin A a sin B b 

since -—r = - , and -r— - = - ; 
sin B b sin C c 

sin J : sin B : sin C :: a : b : c. 



61. Since (Euclid I. 32), the interior angles of a triangle 
are together equal to two right angles, we have 



A + B + C = 180 

. _ sin A a 
Also, 



sin# b 9 

sin 2? b 



sinC c 

And if three of the parts of the triangle be given, the remain- 
ing three parts may be determined by these three equations. 

It is necessary however that one at least of the given parts 
be a side, or we have merely the ratios given which a, b, c 
bear to each other, and their magnitudes cannot be deter- 
mined, because there are but two equations for determining the 
three unknown quantities a. 6, c. 

This also appears from the consideration that by drawing 
lines parallel to the sides of a triangle an indefinite number of 
equiangular triangles can be formed of all possible degrees 
of magnitude. 

62. There is however one case, commonly called " the 
ambiguous case" in which the equations of the last article are 
not sufficient to determine the triangle when three of the parts 
are given. 



42 



If two sides be given, and an angle opposite to one of them 
(a, b, A), the triangle can be determined only when the side 
opposite to the given angle is the greater of the two given 
sides ; i. e. when a is greater than b. 

The equations of the last article become 

5 + 0=180°- A (i), 

b 
sin Z? = - . sin j4 (ii), 

a ' 

sin C 
c = b.— — — (m). 

sin5 v ' 

If we can determine B from (ii), C and c are known from 
(i) and (iii), and the triangle is determined. Now the sine of 
an angle is equal to the sine of its supplement, and therefore 
there are two angles which satisfy (ii), the one greater and the 
other less than 90°. 

(1) Let a be greater than b ; 

.-. A>B. Eucl. i. 18. 

Now B cannot be greater than 90°, for in that case A + B 
would be greater than 180°, which is impossible, (Eucl. i. 17-); 

.-. B<90°. 
Or the lesser angle which satisfies (ii) is to be taken for the 
value of B. 

(2) Let a be less than b ; 

.-. A<B. 
In this case there is nothing to guide us in taking the value of 
B from (ii), since the only limits, 

B + A< 180°, and B > A, 
may be fulfilled whether B be greater or less than 90°. 

Thus if CB=CB\ it is 
evident that both the triangles 
ABC, AB'C have a,b 9 A of 
the same values; also in this 
case ,,W D 

z A is less than the exterior angle CBB\ or the exte- 
rior angle CB'D ; 




43 

i. e. A is less than CB'A, or CBJ; 
.-. a < b. 
Which agrees with what has just been asserted. 

63. In the solution of triangles there are various artifices 
used, which will be pointed out in each case. We shall now 
prove some formulae of great use in the solution of plane recti- 
lineal figures. 

64. To find the cosine of an angle of a triangle in terms 
of the sides. 

Let ABC be a triangle, and from C draw CD perpendi- 
cular to AB or AB produced either way. 




D 



B B 




Then, figs, l, 2, CB 2 = AC 2 +AB 2 -2AB . AD, (Eucl. ii. 13.) 
fig S, CB 2 = AC 2 +AB°+2AB . AD, (Eucl. n. 12.) 

, AD 
And —— = cos CAD, 
AC 

— cos CAB in figs 1, 2. 
= — cos CAB in fig. 3. 
Therefore in both cases, 

CB 2 = AC 2 + AB 2 -2AB. AC . cos CAB ; 
AC 2 + AB 2 - CB 2 



.-. cos CAB 



Or cos A 



2AC.AB 



b 2 + c 2 - d 
2bc 



44 



65. To shew that cos 



>/m- 



where S = 



a + b + c 



For 1 + cos A = 1 + 



b 2 + c 2 - ct 
26c 



b 2 + 2bc + c 2 -a" 

26c 
(b + cf - a 2 

2bc 
(6 + c + a) (b + c - a) 
2bc 



Now, 1 +cos A = 2 (cos — I , Art. 35. (2). 

And S - a = ^ (a + b + c) - « = 1 (6 + c - a) ; 
/ -4\ 2 (6 + c + a) (6 + c - a) 



. . A 1 LU^ — _ 

\ 2/ 26c 




2S .2(S -a) 




2bc 




--ivK'i 




. A /f(S - b) (S - 
66. To sAcw £Aa£ sin — - V/ < , 

2 v ^ be 


-c) 


By Art. 35. (3), we have 




/ . Ay j b 2 + c 2 -a 2 

2 sin — = 1 - cos A - 1 ; 

V 2/ 26c 




a 2 -(b- cf 




2bc 




(a + c -b) (a + b — c) 





! 



26c 






45 

Now S - b = ± (a + 6 + c) - 6 = -| (a + e - 6), 
and *9 - c = 1 (a + 6 - c) ; 

[sin - = — '—— s - J ~ ; 

2 2bc 



■^-V{™^}. 



The positive sign of the root must be taken, because A being 

A 
an angle of a triangle is less than 180°, and therefore sin — is 

a positive quantity. 

Cor. l. (Sin A) 2 

= 1 - (cos A) 2 = (l + cos A) (l - cos A) 



_ J b 2 + c 2 - a 2 \ i b 2 + c 2 - a 2 \ 
= [ + 2b~c J { 2bc J 



46V 

1 



. (b + c - a) (b + c + a) (a + b - c) (a + c - b) 



- . 2 S . 2 (S - a) . 2 (S - b) . 2 (S - c) 



4b 2 C 



= ^-S(S-a)(S-b)(S-c); 
: sin A = ?-*/{8 .{S-a) (S - b) (S - c)} 



. A 

sin — 

A 2 

Cor. 2. Tan — = 

2 A 

cos — 

2 



/l (S-b)(S-c) ) 



67- We will now explain the meaning of the double sign 
in the second member of the equation 



A 

cos — 

2 



- V{^} 



46 

Since cos (4rc . 90° ± A) = cos A, Art. 24. (i) ; 

b 2 + c 2 - a 2 

.-. cos (4rc . 90° ±A) = : 

' 2bc 

Whence we get, by following the steps of Art. 65, 

Now the first member of this equation ought to furnish us with 
two sets of angles whose cosines are of the same magnitude but 
of different signs. Let us see if this be the case. 

First. Let n be even, and = 2 m. 

Then cos ( 2 n . 90° ± - J = cos ( 4m . 90° ± — ) , 

which gives the series of angles 

A A A A 

± 360°+-, 2.360°±— , 3.360°±_, 

2 2 2 2 

whose cosines are all of the same magnitude as the cosine of 

A A 

± — , or oi + — . 

2 2 

Second. Let n be odd, and = 2wi + l. 

Then cos Un . 90° =*=-)= cos U™ . 90° + ( 180° ± - ) 1 , 
which gives the series of angles, 

180° ±-, 360°+ (l80°±- > ) , 2.360°+ flSO ^-^ , 



whose cosines are all of the same magnitude as the cosine of 

A\ A 

cos — . 

2 



180° ± — , which = cos ( 180° | 

2 V 2/ 



Wherefore the first member of the equation does furnish us 
with two sets of angles whose cosines are of the same magnitudes 



but of different signs. 



47 



Cor. J, being an angle of a triangle, is less than 180°; 



68. In like manner it may be shewn that the two values 
of sin — are comprehended under the forms 



sin f 4m . 90° ± — ] , which = sin f ± 

sin| 4ra.Q0 + (l80°±— J 1, which = sin ( 18° ± — j =-sin ( ± — J . 



And A being less than 180°, we have 
(S-b)(S-c) 



■ A t /f ( 



sin 



be )' 

(_A\ ./l (S-b)(S-c) \ 

I !/" V 1 S.(S-a) /' 



Similarly, tan — = + 



tan 



And sin^=+— +/{S.(S-a) (S -b) (S - c)} 
be 



sin (- A) = - — y/{S.{S -a) (S -b) (S-c)}. 
We shall now proceed to the solution of triangles. 



48 

I. ON THE SOLUTION OF RIGHT-ANGLED TRIANGLES. 

69. The right angle, a side, and another part being 
given, to determine the remaining parts. 

Let ABC be a right-angled triangle, C being 
the right angle. 

(1) Let c and A be the other given quantities. 

mi b 

Then - = cos A ; 



c a c 

.-. \ 10 b = l l0 c + L cos A - 10 ; which determines 6. 

Again, - = sin A ; 

.-. 1 10 « = l 10 c + Z sin J - 10; which determines a. 

Again, z 5 = 90° - A ; which determines B. 

Had J5 been the given angle, the method of solution would have 
been the same. 

(2) Let A and b be given. 

c 
Then - = sec A ; 
b 

.-. l 10 c = 1 10 6 + L sec J - 10; which determines c. 

a 
Again, - = tan A ; 

.-. l lo a = 1 10 6 + L tan ^ - 10 ; which determines «. 
Again, B = 90° - A ; which determines 5. 

(3) Le£ A awd a be given. 

a 
Then - = tan A ; 
ft 

a . 

.-. b = , or = a . cot A. 

tan A 

And l lo b = l lQ a - L tan A + 10, 

or = / 10 a + Z, cot A - 10. 



49 



Again, - = sin A ; 



.-. c - - — - , or = a . cosec A. 
sin ^4 

And l lo c = l lo a - L sin A + 10, 
or = l lo a + L cosec A - 10. 
Again, B = 90° - J. 

(4) Z,e£ a, b 6e given. 

Then, tan A = - ; 
b 

.-. Z, tan A = Z 10 « - l m h + 10. 
Again, ^ = 90° - A. 

Again, - = sec A ; 
o 

.-. ^ 10 c = l lQ b + L sec A - 10. 



We might determine c from c = v'C 6 * 2 + ^)? DUt tne opera- 
tion is tedious, particularly if a and 6 be large. 

(5) Le£ c «wrf a 6e given. 

Then, sin ^ = - ; 

c 

.-. L sin J = £ io a - / 10 c + 10. 

Again, - = cos A ; 
c 

.\ / ]0 6 = Z 10 c + Zcos^i — 10. 
Or b 2 - c 2 - a 2 = (c + a) (c - a) ; 

•'• W = J {^io0 + a) + Jio(c ~ «)}; 
which determines 6 without previously finding X 

G 



50 



70. Different methods must be used in different cases 
to determine the unknown quantities ; for what is said in 
Appendix 11. 15. must always be carefully borne in mind, 
and such a formula selected as will give the result with the 
greatest practical degree of accuracy. 

Thus in the last case, if b be very small, the angle A is 
nearly a right angle, and the variation of sin A (i. e. cos A 
sin SA -IsinJ (sinSJ) 2 . Art. 51. Cor. 1.) is very inconsider- 
able for a small given increment ($A) of the angle. In this 
case therefore the value of A cannot be determined from the 
Tables with great accuracy. (Apps. n. 15. ii. ; in. 7-) 



The better way of determining A in such a case will be 
to fii 
its cosine. 



first to find the value of 6, and then to determine A from 



Thus cos A = - ; 
c 

.". Leo?, A = l w b — l lQ c + 10, 

= ho y/ 2 - a 2 ) - l lo c+ 10, 

= J {ho (c+ a) + l 10 (c - a)} + 10 - l lo c. 

71. Example. Having given c = 365*1, and a = 348*3, 
required A. 

(The Logarithms employed in the Examples which we 
shall give, are (generally) to be found in the three pages of 
Logarithms, which are subjoined to Appendices 1. and 11.) 

Here c + a = 713*4, 

c-a= 16*8. 

Performing the operations indicated in the last line of the 
last Article, we have 



51 

l l0 713-4 = 2*8533331 
/ ]0 16-8 = 1-2253093 



2)4-0786424 

2 ! 0393212 
10 



12-0393212 
l l0 365-1 = 2-5624118 



9*4769094 
And L cos 72°, 33' = 9*4769380 See App. in. 8. Ex. 4. 

Difference = 286 

Now diff. for i" is in this case 67*016 ; 

and = 4-267 = 4*27 nearly, 

67-016 J 

Therefore A = 72°, 33', 4"-27- 

II. ON THE SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 

72. Let two angles and the side between them be given. 
(A, C, b.) 

Since A + B + C = 180°, 

.-. B = 180° - {A + C) ; which determines B. 

a • , sin ^ 

Again, a = b - — — - ; 

.-. Z 10 a = Z 10 6 + L sin ^ - L sin 5; which determines a. 

Again, c = 6 -r—- ; 
sin Z? 

.•. / 10 c = / 10 6 + Z, sin C - Z, sin 2? ; which determines c. 



52 



73. Let two angles and a side opposite to one of them 
be given. (A, C, a.) 

Then, B = 180° - {A + C). 

a • *. sirijB 

Again, 6 = a — ; 

sin J 
.*. Z 10 6 = l 10 a + L sin B — L sin X 

sin C 

Again, c = a — — - ; 
sm A 

.-. l 1Q c = l 10 a + L sin C — L sin X 

74. Let two sides and the angle included betiveen them 



he 


given. 


(c, 


A,b.) 












First, 


to 


determine B anc 


C. 














B + C= 180°- A, 














.'. i(*+C) = 


i(l80° 


-A). 








Also, 


b 
c 


sin B 

sin C ' 

6 

1 

c 

- + 1 
c 


sin 5 
sin C 


- 1 








sin i? 
sin C 


-t 

+ 1 






b 
b 


— c 

+ c 


sin B — sin C 


tanl 
tanl 


(B- 
(B + 


C) 


A r 




sin B + sin C 


xiT 




And 


tan 


1 ( £ + C) - tan 


i (180° 


-J) 


= cot 


A 

~2 ; 






•'• 


tan £ (5 - C) = 


6-c J 
cot — ; 

b + c 2 







A 

L tan 1 (5 - C) = / 10 (6 - c) - Z 10 (6 + c) + Z,cot — . (2). 



9. 



\(B — C) being thus determined, and 1 (JB + C) being known, 

B = |(fi + C) + l(fi-C) 1 

, , > are known ; 

and C = i(5 + C)-£(£-C) J 

sini , 
And « = c -r—zz determines a. 
sm C 



53 



75. The side a can be determined without previously 
determining B and C. 

. b 2 + c 2 - or 
For, cos A = 



2bc 

.-. a 2 = b 2 + c 2 — 9,bc cos A 



= b 2 +c 2 -2bc\l -2 (sin-") 1 . Art. 35. (3). 

/ A\ 2 
- (b - c) 2 + 4>bc I sin — j 

of 46c / . J\ 2 1 



(6 



2\/bc . A s 



-«)■{- (¥S*l)i- 



Now, sin — may be of any magnitude and sign, 

b — 2 2 f 

and therefore there is some angle whose tangent is equal to 
this quantity. Let be this angle, 

Then tan0 = ^-^sin- (l). 

b-c 2 w 

and a 2 = (6 - cf { 1 + (tan 0) 2 } 
= (6-c) 2 .(sec0) 2 
.-. a = (6-c)sec0 (2). 

From (1), Ltanfl = l 1Q 2 + ^l 10 b + ^l l0 c-l 10 (b-c) + Zsin — ; 
which determines 0. 

From (2) 3 l l0 a = / 10 (6-c) -f Z, sec - 10; which determines a. 



54 



76. If b=c nearly, b—c is very small, and — — — sin—, 

A A 

the value of tan 0, is very large (unless — and therefore sin — 

be very small), and therefore = 90° nearly. 

Now if from its tangent we have to find an angle which 
does not consist of a certain number of degrees and minutes 
exactly, the additional seconds are generally determined by 
the principle that the increment of the tabular logarithm varies 
as the increment of the angle. But when the angle is equal 
to (2n + 1)90° nearly, this principle does not obtain for the 
tangent, App. n. 11; and therefore cannot in this case be 
determined with an accuracy sufficient to enable us, from the 
equation a = (6 — c) sec 0, to find a exactly to seconds. 

In the case therefore when c is nearly equal to b, and 
A is not a very small angle, the following method will give 
a with more exactness than the last Article does. 

a 2 = b 2 + c 2 — 2bc cos A 

= b 2 + c 2 -2bc.<2 (cos- ) -li Art. 35. (2). 

/ A\ 2 
= (b 2 + c 2 + 2bc) - 4>bc I cos— J 

Now (\/b — vc) 2 or b — 2 y/bc + c is necessarily a 
positive quantity; therefore the positive part of it is greater 

/r- 2\/bc . 

than the negative part, or b+c > 2 \/ be ; .-. — is 

b + c 

2\/bc A 

fractional, and a fortiori cos — is fractional. 

J b + c 2 



55 



Let therefore be such an angle that 

. 2\/bc A ' 

sin v = cos — (1). 

b + c 2 

Then, a 2 =(b + c) 2 {l - (sin 0) 2 } 

.-. a = (b + c) cos0 (2). 

From (1), Zsin0=Z lo 2+ lZ 10 6 + l/ 10 c-Z 10 (6 + c) + Lcos — . 

(2), Z 10 a = Z 10 (b + c) + L cos - 10 ; 
which give # and « respectively. 

77* -^^ ^° «*rfc« be given and an angle opposite to 
one of them, (a, b, A.) 

We have shewn in Art. 62, that the solution is ambiguous, 
unless a be greater than 6. If a be greater than 6, we have 

sin B = - sin A ; 

Where B is an angle less than 90°. 

Also C=180°-(J + B). 

And c = a . 

sin A 

78. Let all the sides he given, (a, b, c.) 

Now sin A = ^- o/{S(S - a) (S - b) (S - c)}, 



. A 
sm — 

2 



cos 



tan 



A _ /\ (S-b)(S-c) ) 
2 " V \ S(S-a) J" 




56 

If A be nearly 90°, the first formula will not give the 
value of A very exactly, because the increment of sin A does 
not in that case vary as the increment of A 9 and it is also 
very small; App. n, 11. and in. 7; — in this case any one of 
the three last forms may be used. And the greater advantage 
will arise from using the second or the third form according as 

A . A . , . A . _ 

cos — or sin — is the greater, l. e. as — is less or greater 

than 45°. App. it. 16. The fourth is applicable in all cases 

j 
except those in which — is nearly 90°. App. n. 11. and in. 7- 



79- Examples. 1. If BC be a perpen- 
dicular object standing on a horizontal plane, its 
height may be determined by measuring in that 
plane a line AC, which is called a base, and ob- 
serving the angle BA C with a proper instrument. 

For BC = AC .tan B AC 
.'. l l0 BC = l w AC + L tan BAC - 10. 

2. If it be not possible to come to the foot of the object, 
let a base AD be measured, such that the points D, A, C may 
be in the same line, and let the angles BDA, BAC he observed. 

Here we have given two. angles and a 
side of the A BAD. By determining the 
side BA we can, from the right-angled 
triangle BAC, find the height of BC. 

_. BA sin BDA sin BDA 

Thus, = — — = 

AD sin DBA sin (BAC - BDA) 

A A r>n o a ■ n a^ An sm BDA. sin BAC 

And BC = BA . sin BAC = AD . — ; 

sin (BAC - BDA) 

•.l }0 BC=l ]0 AD+LsmBDA + LsmBAC-Lsin(BAC-BDA)-Hh 




57 



3. If D be not in the line AC, the height BC can yet 
be determined by taking proper observations. 



At A let the angles BAC and BAD 
be observed, and at D the angle BDA. 

Then in the A BDA we have the 
angles BDA, BAD and the side AD D 
given. If then BA be determined, we can 
find BC from the right-angled triangle 
BAC. 




Thus, 



BA sin BDA 



sin #IM 



JZ> sin DBA sin { 1 80° - {BDA + BAD) } 
sin BDA 

~ ^JbdaVbad)' 

x i r>n ^4 • t> a^ a t. sin BAC .sin BDA 
And BC = BA. sin jBJC = AD . 



sin (BDA + £.4Z>) 



l l0 BC=l l0 AD+Ls'mBAC+LsinBDA-L$in(BDA+BAD)-W. 



Cor. It is evident that it is of no consequence to the 
determination of AB whether D be in the same horizontal 
plane with A and C or not. Hence it appears that to deter- 
mine the height of an object BC above the horizontal plane 
passing through A, we must measure a straight base AD 
in any direction from A, and at A observe the angles BAC, 
BAD, and at D the angle BDA. 



By working many examples of this kind the student will 
become familiar with the management of Trigonometrical for- 
mulae. We shall add two Examples a little more difficult 
than the preceding, before we leave this part of the subject. 

H 




58 

4. Required the error in height arising from 
a small given error in an observation of the angle 
in Example 1. 

Let BC = h, AC = a, z BAC = A. 

A 

Let Sh be the error in height produced by an error S A 
of the observed angle. 

Then h — a tan A, 

h + Sh = a tan (A + $A) ; 

.-. Sh = a {tan (J + §A) - tan A] 

(sin(A + $A) sinA\ 
\cos ( A + $ A) cos J / 

sin (A + $A) cos J - cos (A + S-4) sin A 
cos (^4 + S-4) cos A 

sin {(J + 31) -J } 
cos (A + $A) cos J 

= a ~, 7To • Since cos A = cos (^4 + $A) nearly. 

( cosJ ) Art. 52. 

Cor. Hence we may determine when the error in height, 
arising from a small given error in the observed angle, will 
be the least. 

~ sin£^ h sin §A 

' (cos A) 2 tan J (cos A) 2 
hsm§A Zh.smhA 



sin J . cos A 2 sin A . cos J 

2& sin 3 J 



sin 2 J 
Now h and §A being constant, this expression is the least 
when sin 2 A is the greatest, which is when 2^ = 90° or A = 4,5*. 

The observer therefore ought to move along AC till 
z BAC = 45°, and then by measuring AC he will determine 
CB, which is in this case equal to AC, with the least chance 
of error. 



59 



5. The angular distance at noon of a kite's shadow from 
the north is /3, the wind blowing at an angle a from the south. 
The wind suddenly changed to a from the south, and the 
shadow to /3' from the north, and the kite was raised as much 
above 45° as it had been below that elevation. Required 
the angles of the altitudes of the kite and the Sun. 

Let B be the place of the spectator, 
N the north point, K the kite, KG per- 
pendicular to the horizontal plane through 
B, S the shadow of K on that plane. 
Then the hour being noon, the shadow 
of the kite is due north of it, and there- 
fore SG is parallel to NB. Also z KSG 
is the Sun's elevation above the horizon, 
and z KBG that of the kite, z GBN is 
the angular distance of BGK from the 
north ; therefore BG is the direction of the wind. 

Let £KBG = 4,5°-(j>, z KSG = 0. 

LNBG=LgBn = a, NBS = )3, .: zSBG = a - fi. 

Now — — = cot KBG = cot (45° - <j>) ; and — = tan ; 




KG 

, n v BG SWBSG 

,\ tan . cot (45° - 0) = = — 



SG 



sin SBN 
sin SBG 

tan0 



SG sin SBG 

, since SG is parallel to BN; 

sin/3 



tan (45° - 0) sin (a - )3) 



(1). 



Now on the change of wind, 6 remains the same, while 
a, /3, 45° — 0, become a, /3', 45° + ; whence we have in 
the same manner, 

tan0 sin/3' , x 

tan (45° + 0) sin («'- /3') V y 

And from (1) and (2) and cj) may in the following 
manner be determined. 



60 



And 



Dividing (l) by (2), we have 

tan (45° + $) _ sin j3 sin (a' - /3') 
tan (45° - <jf>) " sin/3' * sin (a - /3) ' 

tan(45°-(/)) = COt(45 ^ 

sin j3 sin (a - /3') 



.-. {tan(45°+0)} 2 =^-g 7 . -^ ^ 

^ 7 3 sin /3 sin (a - /3) 



•(i) 



And from (2), tan0 



sin/3' 



— tan (45° + <p) 



sin («' - /3') 

sin/3' /sin/3 sin (</ - /3') 

~ sin (a - /3') ' ^ sh^/3' ' sin (a - /3) 

/ sin /3. sin /3' ... 

= V sin (a - j3) sin (a' - /3') " ,(ll) ' 

From (i), Z, tan (45° + 0) = | {L sin /3 + £ sin (a - ft) 
- L sin /3' - Z, sin (a - /3)} -10. 
(ii), L tan = ^ { Z, sin /3 4- £ sin /3' - L sin (a - /3) 
-Z,sin( a '-/3')}-10. 
Whence <£ and are determined. 

80. To find the radii of the circles described within and 
about a regular polygon of any number of sides. 

Let AB be a side of a regular polygon of n sides. 

Then since the polygon is regular it 
may have a circle inscribed in and about 
it, and each of the sides will subtend the 
same angle at the common center C of these 
circles. 




61 

Draw CD perpendicular to AB. 

Then AD=DB, and CD is the radius of the inscribed circle. 

Let r=CD, and B = AC. 
Now the sum of all the angles which the sides subtend at C 
= n.L ACB = 360° ; 

n 



And— - tan-; 

C 180^ 
.-. r=CD= ADcot-=±ABcot 

2 ^ rc 



,(1). 



Again, R = AC = 



AD 



= -£- -=^JBcosec ...(2). 

sin JCZ> . 180° 2 ro W 



sin 



?z 



81. To find the area of a triangle, the sides being given- 




Draw CD perpendicular to AB or AB produced either 



way. 



Then area of triangle ABC 

= \ AB. CD = i>AB. AC. sin CAB 

= ±cb.^- v /{S(S-a)(S-b)(S-c)}. Art. 66. Cor. 

- x/{S(S - a) (S - b) (S - c)|. 



62 

sin B . sin C 



82. The area of the triangle also = 1 a 2 



sm(B + C) 



For area = \AB. CD 



= \. AB. BC. sin B 

= ±BC.^.BC.sinB 
sm A 

, 2 sin B . sin C 
= 1 * a sin (5 + C) ; 
Since sin ^4 = sin {l80° - (B + C)} = sin (5 + C). 

83. To find the radii of the circles described in and about 
a triangle whose sides are given. 

Let the lines bisecting the angles A and 
B meet in O, and from O draw OD, OE, 
OF perpendiculars to the sides. 

Then, Euclid iv. 4, O is the center of 
the inscribed circle, and r its radius = OD A 
= OE= OF. 

Now, area of A ABC = A AOB + A BOC + A CO A ; 

.-. ^/{S(S-a)(S-b)(S-c)} -r.l+r.j+rl-r.Si 




.-. r == \/ 



(# - a) (S - b) (S - c) 



Again ; bisect the sides of the triangle 
in Z), E, F, and draw perpendiculars which 
will meet in a point O which is the center A 
of the circumscribed circle; and R its radius 
= OA = OB = OC. Euclid iv. 5. 

Area of ABC = \AC . AB. sin A, Art. 81. 

And A=\ BOC, Euclid in. 20 ; 




63 



.-. sin A = sin \ BOC = sin BOE = — = ^ ; 
• '. %-y { S (S - a) (S - b) (S - c) } = ^ 5 



.-. # = 



a&e 



♦ V^^-^^-iXS-f)} 



••(2)- 



84. To ^wrf the area of a quadrilateral figure whose 
opposite angles are supplements to each other. 

Let ABCD be the quadrilateral. 

Let AB=a, BC = b, CD = c, DA = d; 

Join AC. a 

Area ABCD = A ABC + A ADC 

= ^«6 sin 5 + ijcd sin Z> 

= -| (ab + erf) sin J5. 
Since sin D = sin (180° - D) = sin 5. 




Now from A ABC, 

a 2 + b 2 -AC 2 = 2ab cos 5. 
And from A ADC, 

c 2 + d" - AC 2 = 2 erf cos 2) = - 2cd cos i? ; 
.-. by subtracting, 

a 2 + b 2 - c 2 - d 2 = 2 (ab + cd) cos 5 ; 

,. (sin Jgf-1.- (cos B). - 1 - ( y-^-'T 
' v 7 \ 2 (ab + erf) J 

_ 4 (aft + erf) 2 - (a 2 + b 2 - e 2 - rf 2 ) 2 
4 (ao + erf) 2 



64 
And (area ABCDf = 1 (ab _+ cd) 2 . (sin £) 2 
.-. = -1 {4 (ab + erf) 2 - (a 2 + & 2 - c 2 - d 2 ) 2 } 

= — {2(a6+cc/) + (a 2 +6 2 -c 2 -rf 2 )}. {2(ab+cd)-(a 2 +b 2 -c 2 -d 2 )} 

= ^ { (? + h Y - ( c - d f\ • {^ + <0 2 " (« " &)*} 

= — . (a + b+ c-d) (a + b + d - c) (c + d + a - b) (c + d + b - a). 

If S = i> (a + b + c + d), this equation becomes 
Area ABCD = ^ {(S - a) (S - b) (S - c) (S - d)\. 



CHAPTER V. 



ON THE USE OF SUBSIDIARY ANGLES. 



85. Def. An angle which is introduced to assist Trigo- 
nometrical calculation (as in Arts. 75, 76.) is called a sub- 
sidiary angle. 

86. Subsidiary angles are of great use and extensive 
application, particularly in adapting the formulae of Astronomy 
to logarithmic calculation. 

Ex. 1. Having given a, a, and S, to determine 1 in a form 
adapted to logarithmic calculation from the equation. 

Sin a = cos I cos $ cos a + sin I sin § ; 

{Maddy's Astronomy by Hymers 9 Art. 118.) 

The equation may be written, 



sin 



. * / . . , cos d cos a\ 

a = sin d • I sin I + cos I . : — s — 1 

V sin 6 ) 



Now since tangents are of every magnitude and sign, there is 
an angle 0, such that 

sin <h cos S cos a * 

tand), or V = : — ~ — = cot d cos a (1); 

r cos sin a 



• * / • 7 , sin 0\ 
a — sin d I sin I + cos / . — 

\ cos <pj 

(sin Z cos + cos I sin 0) 



sin 

sin£ 



COS0 

sin 3 . 
= -sin (/+ 0); 

COS0 

. ,. , x sin a . cos d> , x 

I 



66 
From (l); L tan <p - L cot $ + L cos a - 10, which gives (p. 
(2), L sin (I + <f>) = L sin a + L cos - Z sin S, 
which gives I + (p, and thence Z. 

Ex. 2. Having given 1, S, S', a, £o determine h m a form 
adapted to logarithmic computation from the equation 

tan / (sin 3 — sin 5') = (cos $' cos a — cos $) cos h — cos ^ sin a sin ^ ; 

(Hymers* Astronomy, Art. 183.) 

Sin S - sin 3' = 2 sin 1 (3 - 3') cos 1 (5 + 3'), Art. 44. (2); 

.-. tan/.2sinl(S-S')cosiL(S + S') 

S' cos a — cos $ 



= cos 



ft, . I cos a cos a - cos d , . , ) 

6 sin a < ^7—; cos h - sin A > 

( cos o sin a J 



Now, since cos $' cos a is less than 1, let be an angle such 
that 

cos0 = cos $' cos a .....(l); 

.*. cos & cos a — cos 3 = cos — cos $ 

= 2 sin 1 (3 - 0) sin | (S + 0), Art. 44. (4) : 

and the proposed equation becomes 

2 tan I . sin 1 (3 - S') cos 1 (3 + 2') 



= cos 



sin a < — kt-. — cos h - sm h ) 

[ cos 6 sm a J 

Again, let be an angle such that 

2sinl(^0)sinl(3 + 0) 
cot0= ^- sp^ — - (2). 

' cos d sm a 



67 

And by the substitution of cot (p, we have 
2 tan I sin \ ($ - <5') cos J (cS + <S') 

= cos S' sin a (cot cos & — sin ^) 

(cos cj) cos & - sin sin h) 
cos (0 + A) ; 



sin 

cos §' sin a 
sin <p 



s-l n 2sin0. tan/. sinl(3-3')cosA(3 + 3') /x 
.\ cos(0 + A)= 21 — ?_A_ ^_ — 2_\ ^...(3). 

' cosd sin a 

From (l) L cos = Z, cos tf + L cos a - 10 ; which gives 0. 

(2) Z cot <p = Z 10 2 + L sin ^ (S - 6) + L sin 1 ($ + 9) 

— L cos S' — L sin a + 10 ; which gives <p. 

(3) L cos (0 -f &) = Z 10 2 + Z, sin cf> + Z, tan I 

+ Z, sin| (S - S') + Lcosl (S+ £') 
— Z, cos S' — L sin a — 10 ; 
Whence h may be found. 



CHAPTER VI. 



ANALYTICAL TRIGONOMETRY. 



87- The circumference of a circle varies as its radius. 

Let P, p be the perimeters of two regular polygons of 
n sides each, which are inscribed in two circles whose radii 
are R> r and circumferences C, c respectively. 





LetX=C-P, w^c-p. 

Now as n increases, P and p increase ; therefore X and 
x are variable quantities dependent on the value of n. 

Let AB, ab be sides of the polygons; C, c the centers of the 
circles. 

360° 

Then I ACS = = lacb, 

n 

and ACB, acb are similar triangles; 

R AB n.AB P C-X 



r ab n . ab p c — oc 
.-. R.c- R.a; = r.C -r.X; 

Now since R.c, r.C are constant, and R . m, r . X are va- 
riable, we must have, Francceur's Pure Mathematics, Art. 167, 

R.c m r.C; 

c r 
.-.—=—, or c oc r. 
C R 



m 



_, „ c 

Cok. 1 Sum k - = a constant quantity 

This constant quantity is always represented by ^ x ; the 
approximate value of x. 5- 141 op... . will be determined here- 
after. 

c 
Lor. S Since - = : - : .. e = i?xr; or xxr represents 

the circumference of a circle whose radius is r. 



If an arc be traced out by a point in the line CB. 

by tchose revolution from the position CA an angle ACB 

is described, the angle ACB way be properly measured by 

arcAB 

the ratio — 

radius Al 

Since in equal circles, and therefore in 
the same circle, angles have the ratio of the 
arcs on which they stand. Euclid vi. ; 

-ACB arc AB 



4 right angles whole circumference ABDA 
AB 




Art. 87. Cot 
2x. LA 

4 rio-ht antrl 

iACB = 



: - CA 

4ricjht angles AB 



:- 

M 4 right angles AB 

- Now s:nc - — is a constant quantitv. increases 

* CA 

or decreases in proportion as the angle ACB increases or 

decreases; therefore — — — is a proper measure of the naff- 
radius r r ^ 

nitude of an angle, and we may say that the angle is equal 

arc 
to 



radius 



Hence also, if 9 = — ^— — . then arc = r& 
radius 



70 

89- We have in the preceding Chapters measured the 

magnitude of an angle by the number of times it contains a 

fixed and definite angle called a degree, and its subdivisions ; — 

for several analytical investigations, however, the circular mea- 

arc 
sure — - — is much more convenient, 
radius 

When this circular measure is used, we shall, generally 
speaking, denote angles by the letters of the Greek Alphabet. 

arc 
90. Having given the circular measure — — — oj an 

angle, to determine how many degrees the angle contains ; 
and conversely. 

Let be the circular measure of the angle which contains 
A degrees. 

circumference _ _ „ - , 

Since - = 2tt, and the circumference subtends 

radius 

four right angles, therefore 2tt is the circular measure of four 
right angles. 

A given angle 

Now == -A— °— =± ; 

360° 4 right angles 2 7r 



n A 7T . 

.-. = 2tt.— = A (1). 

360 180 



And A = 360.— = — .0 (2). 

27T 7T 



Ex. 1. If A = 60, 



/» 6o 

0= .7T = 1 x 3-14159 ... - 104719 

180 3 



71 

Ex. 2. Required the number of degrees subtended by the 
arc which is equal to the radius. 

arc 
In this case = — - — = 1 ; 
radius 

J 180 n 180° 

... j = Q = = 57° '29577 

7T 3-14159- 

60 



17-7462 
60 

44-772 

And the degrees, minutes, and seconds required are 

57°, 17', 44" -77. 

It is sometimes necessary to know the number of seconds 
subtended by an arc which is equal to the radius. 

The number of degrees subtended = 57*29577 

60 



3437-7462 
60 



.-. number of seconds subtended = 206264-772 = 206264-8 nearly, 

91. Four right angles being represented by 2 7r, we have 
from Arts. 22, 23, 24, if the angle A be represented according 
to the circular measure by 0, 

sin = ± sin(2w7r =*= 0), or = ± sin {(2n + l) <k =f 0}> 

tan0 = ± tan(2w<7r±0), or = ±tan \(2n + 1) tt±0}, 

cos0 - cos(27Z7r ±0), or = - cos {(2n + l) tt ± 0}, 

sec = sec (2w7r =*= 0), or = - sec {(2n + l) 7r ± 0} . 



72 

92. If x + - represent 2cos0, then x — will represent 
2 \/-~l sin 0. 

For - (sin 0) 2 = (cos0) 2 - 1 = \Lc + -V- 1 ; 

.-. - 4 (sin Of = x 2 - 2 +i ; 

a? 8 

. l 

.-. 2 v - 1 sin0 = w - ~ . 
x 

Demoivre , s Theorem. 

93. To shew that for any value of m, 

(cos =*= v— 1 sin 0) m = cos tw0 ± \/- 1 sin m0. 



(Cos0±v'-1 sin0).(cos0±A/-l sin0) 

= (cos0) 2 - (sin Of =*= \/~ • 2 sin . cos 0. 

Or, (cos0 ± \/^~l sin 0) 2 = cos 20 ± V'- 1 sin 20. 

Again, (cos =fc \/ - l sin 0) 2 . (cos ± y^ - 1 sin 0) 
(cos 2 ± \/~ sin 2 0) . (cos ± <\/ -~1 sin 0) 
cos 20 . cos - sin 2 . sin =k\/ - 1 (sin 20. cos 0+cos 20 . sin 0) ; 
\ (cos0±^/-l sin 0)* = cos 30 ± >y/~ sin 30. 



73 



Suppose this law to hold for m factors, so that. 

(cos =t \J - 1 sin 6) m - cos mO ± V - 1 sin m0 ; 

.-. (cos0± y/~^l sin6) m+l 

- (cos ra0 ± v - 1 sin ra0) . (cos ± \/- 1 sin 0) 

~ cos mO . cos0 - sin mO .sin0 ±\/ - 1 (sinm0.cos0 + cosm0.sin0) 

= cos (?w + 1) ± v- l s i n (w* + 0* 

If therefore the law hold for m factors, it holds for m '+ 1 
factors ; but we have shewn it to hold when m — 3, it is there- 
fore true when m = 4, and by successive inductions we conclude 
it to be true when the index is any positive integer. 

— / 1 \ m 

Again, (cos ± \/ - 1 sin 0)~ m = ( ;=== | 

5 ' Vcos0±a/-1 sin0/ 

f (cos0) 2 +(sin0) 2 | m 

icos ± V-^l sin 0J ' 

= (cos0=fV /:: ^ sinG)*, 
by actual division. 

= cosm0=F v - 1 sinm0, 

= cos (-mO) ± v -l sin(-m0); 

which proves the theorem for negative indices. 

Again, (cos ± \/- 1 s^ 0)™ = cos m0 ± \/- 1 sin m0. 



But (cos — 0± -\/- * sin — 6Y = cosmO ± \/- * sinm0 

.-. (cos ± \/^l sin 0) ,n = (cos - ± V^L sin - OY ; 

^ n 

.♦. (cos0± v'-l sin0) w = cos— 0± v /^l sin — 0; 



which proves the theorem for fractional indices. 

K 



74 

Cor. By the theorem just proved, 
(coscp ±y-l sin <p>) m - cos m<p± \/ — l sin mcj}), 
m being positive or negative, whole or fractional. 
Let <£> = 2^7r + 0, where p is any integer ; 
.*. cos (f> — cos (Zpir + 0) = cos 0, 
sin = sin (2^7r + 0) - sin 0. 

1. Let the index of (cos =*= y/ — 1 sin 0) be integral, as w; 

Then cos m(p = cos (2mp7r + m0) = cos mQ, 
sin wz0 = sin (Zmpw + ra0) = sinm0. 

m 

2. Let the index be fractional, as — ; 

n 

Then p being an integer must be of the form qn + r 9 
where q is or any integer, and r is or an integer less 
than n ; 

m m m , . 

.-. cos — </> = cos — (2»7r + 0) = cos — «2 (gw. + r) 7r + 0} 

= cos {mq27r + — (Zrw + 0)} 

= cos — . (2r<7r + 0). 
n 

O- -1 1 . *» . . rn . _ x 

Similarly, sin — <£ = sin — (2r7r + 0). 

From these two cases therefore it appears that the theorem 
might have been thus enunciated ; 

If the index be an integer, 

(cosO ±\/- 1 sin0) m = cosra0 ±\/- l smrnO; 

If the index be fractional, 

(cos0=t v / -Tsin0) w =cos-(2r<7r + 0) ±\/-l sin- (2r<n- + 0); 
7 n n 

where r is 0, or any integer less than n. 



75 

It is to be observed that by giving r all values from to 
n - 1, we obtain from the second member of the second equation, 
the n different values of 

m 

(cos0± V - l sin0) w . 
94. If 2 cos 6 be represented byx + - , then 2 cos m0 will 

be represented by x m -f — , and 2y-l sin m by x m . 

For 2 cos = .r + - •> 

x 

And .-. 2 \/~^~i sin = a? - - , Art. 92. 

a? 

.*. cos0 + v - 1 sin0 = a?; 
.-. #'» s= (cos + v- 1 sin 0) m = cos mO + y^- * s i n ^0 ; 
1 1 

# m cos mO + v - l sin wO 
(cos ra0) 2 + (sin mO) 2 



cos mO + \/ - 1 sinm0 
= cos m - v — 1 sin m 0, by actual division ; 

.*. a?" 1 4- — = 2 cosm0, 



And oe m = 2 y/ 

x m 



1 sinm0. 



Cor. By making use of the equations of the corollary to 
Art. 93, we have, 



The index being an integer, 



cosm0 = oc m H ; 

x m 



/ . ^ 1 

2 V - 1 sin mO = at m ; 



The index being a fraction, 



cos -{2rir + Q) = x n + — ; 2\/ - 1 sin - (2r<7r + 0) = co n - — ; 
where r is or any integer less than n 



76 

95. To express any positive integral power of the cosine 
of an angle in terms of the cosines of the multiples of the 
angle. 

Let 2 cos0 = <#+-; .*. 2 cos nO = co n + — . Art. 94. 



Now (2cos0)" = (a?+ -J 



n — I 



ar~' + .. ? . 



w-i i 11 

+ W. . 1-». H (1.) 



2 .2?" 



#»-" a?" 



(^ + i) + »-{ af " , + ^i) ,+ ' ^ 



1. If n be even; the last term of (2), or the (|w + l) th 
term of (1), is 



n 



n—\ n—2 w-|w+l 



^n 



2. If w be odd ; the sum of the two middle terms of (l), viz. 
the {1 (n - 1) + 1 } th and the {1 (n - l) + 2} th , is 

rc-1 ^_l(^_i) + i 
- .... y 

2 



W . 



±(»-l) + l / 1\ 

1 (» - l) v #/ 



Hence (2) becomes 

(2 cosfl)* = 2cosra0 + w. 2 cos(w - 2)0 + ... 



+ < 



n — 1 w - -^ w + l 

w . . . . ^ ; w even. 

2 iw 



w - 1 w - -|(w - 1) + 1 ,- 

w . . . . r-7 : 2 cos 6/ ; n odd. 



2" _I . (cos 0)" = cos n9 + n. cos (« - 2) 6 



iyt J 

+ n . cos (n - 4) + ... 



77 



the last term being 



n - 1 n — An+ 1 



+ ±-n. 



n 



; n even. 



+ n. 



n-1 n-\(n-l) + 1 



cos0; n odd. 



2 *"" i(^-l) 

Thus, if n — 2, we have 2 (cos 0) 2 = cos 20 + 1, 

n = 3, 4 (cos0) 3 = cos 30 + 3cos0, 

?z= 4, 8 (cos0) 4 = cos 40 + 4 cos 20 + 3. 

96. To express any positive integral power of the sine of 
an angle in terms of the sines and cosines of the multiples of 
the angle. 

Let 2 \/-~T sin = at ; .*. 2 \/ - 1 sin n0 = at 71 . 

w at n 

/ f l\ n 

And (2V-1 sin0) M = lot- -j 



» 2 n ~ l n-4 

2 



W-l 1 1 1 , x 

.±n. =?n -± — ...(1.) 

2 a? 7 *" 4 af- 8 a? re v / 



the upper or lower sign being taken as n is even or oddj 

= (ff»±-_) - n . (at n - 2 ± A + n> ^— [o? s - 4 ± — ) -...(2.) 

V W \ at n - 2 ) 2 \ at 71 -*) K } 

1. If w be even, it is of the form 4m or 4m + 2. 
Now if n = 4m, (yTi)*« = i = (- i) 2j » = (_ i)§ 3 

if w = 4m + 2, ( v / ^1l) 4w2+2 = - 1 = (- l) 2w2+1 = (- l)?. 

(n \ th 
The middle, or I - + 1 J , term of the expansion (l) is, 



i n . 



n-1 



n — -kn + 1 



n 



78 



— hi) , term is an odd 

or even term, i. e. as — is even or odd. Wherefore it has the 

2 

n 

same sign as (- l) 2 has ; and n being even, (2) becomes 
(- l) 2 2" (sin 0) n = 2 (cos nO) - n . 2 cos {n - 2) 

+ n . . 2 cos ( n - 4) - ... 

2 

...+ (-l) 2 rc.— — - ... f ; 

;'••. (- 1) § 2 n ~ l . (sin 0) n = cos nO -n cos (w - 2) 

** - 1 • X /l 

+ w . cos (n - 4) - ... 

2 

... + -1/k.-— ... f . 



2. If n be odd, it is of the form 4m +1 or 4m + 3. 

Now if n = 4m + 1 ; (y^) 4 " 1 * 1 = (a/-^)*" <\/^T 

= ( _ !)- -A /rT = (- i)^ . v^T; 

If rc - 4m + 3 ; (v/rT) 4, " +3 = (V^T) 4 '" 4 - 2 . a/^~1 

= (-i) 2w+1 .V r ^i = (- 1)^.\ " 



V - 1 : 

and the last binomial in (2) is 

n-\ n - ±(n-l) + l ( 1\ 

± w • f-^ . [x . 

2 i(^-l) V W 

+ or — according as there is an odd or an even number of 
binomials in (2), i. e. as 1 (n + 1) is odd or even; the sign 

n+l n-\ 

therefore of this term is the same as that of (-1) 2 ~,or(-l) 2 , 
and (2) becomes, 



79 



(- 1)~^~ 2" \Z~^1 . (sin ey 
- 2<\/ '- 1 sinw0- n.2\/- 1 sin (w - 2) + ... 

+ (-1) 2 n. ... f^ ^ .2<v/-lsin0; 



n - 1 
= sin rc0 - rc . sin (n - 2) + n . . sin (n - 4>) + ... 

... + (-l) »» „. fi ^ sin0. 

2 i(n-l) 

97- ^o expand cos n^iw terms of cos alone, n being a 
positive integer. 

1 ! 

Let 2 cos = a + - ; .-.2 cos nQ — a n -\ . 

a a n 

Now, (1 — a x) f J J = l - f a + - ) a? + a? 2 

= 1 — x (b — x), 

if 6 = a + -=2 cos ; 
a 

Then Z e (1 - aw) + ill --) = l e {l - x (b - x)}. 

Expanding these logarithms, Appendix I. 14. iii, and changing 
the signs of both sides of the equation, we have 



ax + -La 2 a? 2 + ... + - a n x n + ., 
n 



x , or 
— u A — 



1 x n 



+ - + 



2 „2 



+ ... + 



a ~ a* n a 

n.n-2 



+ ... 



} = x(b-x) + — (b -xf + 



+ (6 - *)«- 2 + - — (P - w)"' 1 + - Q> ~ X T + 



80 

And equating the coefficients of w n , 

i / l \ 1 x n 

— ( a » + — ) = — b n {which is the first coefficient of — (b — xY\ 

n \ a n ) n ( n } 

1 w n ~ l 
. (n - 1) b n ~ 2 {the second coefficient of (b - atY' 1 } 

+ . i L± ' b n ~± {the third coefficient of — (b-ocY~ 2 \ 

n-2 1.2 L n-2 K J } 



The general term being the coefficient of that term of 

x n ~ r 

.{b-x) n -% 

n- r 

which involves oo n ;— and this being that term which involves ot? 
in the expansion (6 - w) n ~ r must be the (r + l) th in that expan- 
sion ; — also this term is + or — as r is even or odd ; 

Therefore the general term is 

1 (n — r) . in — r — l)...(n - 2r + l) , _ 

/ 1 \r _> ' V / V ' lM—2r 

n-r' 1.2. 3...r 

Or, (-!)■ >- r -'> ( w - a '"+ 1 )y-.r. 

v ' 1.2.3...r 

Writing 2 cos n for a n + — , and 2 cos for b, 
the equation becomes, when both sides are multiplied by n 9 

71 — 3 

2 cos nO = (2 cos 0) n - n (2 cos 0) w_2 + n . . (2 cos 0)"- 4 - ... 

the general term being 

(_ , )r „ 0L-r-l)...(n-ar + l) (2cos0) „- 2 , 
v ' 1.2.S...r 



81 

In determining the series from the general term by giving 
to r different values, r must be taken of every positive integral 

n 

value not greater than — : for since none but positive integral 

powers of (6 — x) appear, n — 2 r the index of 6 in the general 

term must be a positive integer, or r must not be greater 

i n 
than — . 

2 

In the proof of this Article the expansion of l e (l + x) is 
used, in which it might be shewn that none but positive and 
integral powers can enter. This method of proof therefore 
can only be applied to those cases in which n is a positive 
integer. In fact, it can be demonstrated that cos n$ cannot 
be expressed in a series of this form when n is negative or 
fractional. 



98. To express cosn0, n being a positive integer, in 
a series ascending by integral powers of cos 0. 

In the last article cos n6 has been expressed in a series of 
descending integral powers of cos : by means of the general 
term there found, cos n 6 may be expressed in a series of ascend- 
ing integral powers of cos 0. 

1. Let n be even. 

Since the limitation of the value of r is that it shall not 

, , n n n n 

be greater than -, tor r write - , l, 2, ...3, 2, 1, 

2 2 2 2 

successivelv. 



The number of terms will be ^n + 1 ; and since r is di- 
minished successively by unity, the terms are altertiately 
positive and negative ; the first term being of the same 
sign as (- if ; 

L 



82 

.-. 2 cos nO 



= (- N • - — — \ ( 2 cos 0) 

1 1 . 2 ... -g- 7i 

- w {n-(i«-i)-i}.{Mi"-D-g?-H(i«-i)+l} , 2 cos fl y 
1.2. 3. ..(1^-1) 



{w-( |w-2)-l}.{w-(^w-2)-2}... fo-2(^w-2) + l} 
1.2. 3. ..(1^-2) 

| fn.(£n-l).(£n-2)...3.2.1 



/TV 4 I 

+ W .J 12 i-^ * ' — L__< 1^ £ > (2cos#) 4 -...> 

1 .2.S...(lw-2) 

f - {<n. . (Xr n. — 1 , * (\- vl — 9\ _ . . a .9 . 1 



1 .2...|W 



n.\nA\n- l)...4.3 , 

8 12— i— (2 cos 0) 2 

1.2...(i»-l) / 



n . (| w + 1) . \ n . (1 n - l) ... 6 . 5 
1.2. 3.. .(1^-2) 



(2 cos 



ey-...} 



-i/f*-^- («.«•«■ 



n.(l, + 1 ).X M .(i^-l) p | 

1.2.3.4 ' J ' 



+ — , n a A — - (2 cos 

and dividing by 2, 

- n 2 n 2 (ri z — 9?} 

cos no = (- 1) 2 5 1 (cos ey + — ^ ; (cos ay 

v ' * 1.2 v ' 1 .2.3.4 v 

1 .2.3.4.5.6 V ' > 

2. Let n be odd ; r must not be greater than i n, 
.-. r may = 1 (w — l), the integer next less than ^n. The 
number of terms is l(w+l), and the terms are alternately 
positive and negative. 






83 



For r writing 1 (n - l), 1(^-3), l(rc-5)...3, 2, 1, 0, 
successively, we have as before, 

2cosw0 = 

\ 1.2. 3.. .1(^-1) V f 

n . {n - \ {n - 3) - 1 } ... 6 . 5 . 4 

(2 cos 0) 3 



1 .2. 3.. .1(^-3) 

/ ^ 1 Vr 7 ' 6 (^os^ 5 ...|. 



ft ,{ B ,|( )t .5)-i}„,7.6 /< 
1 .2.3... ±(n - 5) 



And reducing the terms in the same manner as in the last case 
when n was even, we have 

cos nO = 
V J l 1.2.3 V J 1.2.3.4.5 V ' 

_,.(rf-iO.(w-^).(tf-tf) 

1.2.3.4.5.6.7 * 



99- Having given tan 0, to find tanw0. 
Cos ?a0 + v^ — 1 sin w# 
= (cos0 + \/- 1 sin0) w 
= (cos 0) u (1 + V^ tan 0) n 

= (cos0) w . {l + ^ • \/-^ tan0-n. (tan 6)* 

-n.^/-^^—(Uney + ...} (1.) 



84 



And equating the possible and impossible parts of this equation, 

71 — 1 

cos nO = (cos Of {l - n . (tan 0) 2 



n — In — 2n — 3, „ . 
+ n.~^-.— r .^-(t a ney-...} (2.) 



sinnO = (cos Of {n tan -n . -——.—— (tan0) 3 +...} (3.) 

At O 



n - 1 n - 2 . 

„ ntmO-n. . (tan0) 3 +... 

. sin^0 2 3 

.*. tan nO - 



cos nO n - 1 

1 - n . (tan 0) 2 + 



Cor. If n be a positive integer, the series (l) will termi- 
nate, and the last term is (v - 1 tan 0) n . 

1. When n is even, it is of the form 4 m or 4m + 2 ; 
If n = 4m, (\/^T tan 0) w = (<v/^T) 4w . (tan 0) M = (tan 6) n ; 

If w = 4m + 2, (a/^ tan 0) w =(v/^T) 4w+2 . (tan 0) M = -(tan 0) M , 

n 

therefore, in both cases the last term is (- l) 2 (tan0) w , and 

ntanO-n. .— (tan0) 3 +... + (-l) 2 7z(tan0) n " 1 

tan n = . 

rft J If 

l-w. (tan0) 2 + (-l) 2 (tan0)» 

2. When n is odd, it is of the form 4m + l or 4m + 3 : 
If n = 4 m + 1, (>/^~T tan 0)" = x/^l (tan 0)" ; 

If rc = 4m + 3, (v^T tan 0)" = - \/-^l (tan 0)' 1 ; 



85 



Therefore, in both cases the last term is (-1) 2 (tan 0) n v -l, 
7i— 1 n—2 



ntanO-n. 



(tan0) 3 +... + (-l) 4 (tan0)» 



and tanw$= 



n-\ 



«— 3 



l_^.___(tan^) 2 + +(-1) 2 n(tan0) n ~ 1 



*~~ ,.„ „ . sm # , tan# 

100. When is = 0, — — - = 1, a»rf — — - = 1. 
8 

Let JjB be the arc of a circle whose center is C and radius is AC. 

Let z JC5 = ; and let AT touch the circle at A. 

From B draw BN perpendicular, and 
BM parallel, to AC. 

Then MN is a rectangle, and 
AM = BN, MB = AN. 




N A 



Now JZ? > jBiV and < Jilf + MB ; (Legendre's Geometry) ; 



AB BN _ .RftT JiV 



or, < sin and < sin + versin ; 



, 1 - cos 

> 1 and < 1 + 



sin 



sin# 



And 



1 — cos 
sin# 



. ov . 

2 



2 I sin - 1 sin 



. 

2 sin - . cos - cos 

2 2 



- = tan - , 

2' 





— — - > 1 and < 1 + tan - < 
sin# 2 



86 



6 

Since then — — - always lies between 1 and 1 + tan - , 
sin 2 

Q 

and tan - - when - ; 

2 

1, or — — = 1, when = 0. (l.) 



sin0 

tan sin 1 



Also, 



' cos ' 



and when = 0, — — — = 1 , and cos = 1; 


tan _ , , x 

.\ = 1 in this case. (2.) 



101. To shew that cos a = 1 — 

and sin a = a — 



a 2 

+ 

1 .2 


a 4 


1.2.3,4 


a 3 


a 5 

+ 



1.2.3 1.2.3.4.5 
Since, Art. 99- (2), 

n — l, '• n—\ n-2 n—3 



cosn 



0=(cos0) tt {l-rc-^— (tan0) 2 +rc.— . — - . (tan0) 4 -...}, 



let n = - ; 
a 
cos a = (cos 0) n { 1 - - . (tan 0) 2 



a a a 

+ -. . . (tan0) 4 -...( 

2 3 4 V ' * 



, ^ c a-0/tan0\* a-0 a-20 «-30/tan0\' 



87 

Now this is true whatever be the value of 9 ; it is therefore 

true when 0=0, in which ease = 1, and cos 0=1, Art. 100; 

6 

a 2 a 4 , x 

.-. cosa = l 1 (1.) 

1.2 1.2.3.4 

By a similar substitution in the expression, Art. 99. (3), 
sin nO = (cos 6) n {n tan 9 - n (tan 9f + ... } , 

we obtain 

a 3 a 5 . . 

sin a = a h (2.) 

1.2.3 1.2.3.4.5 

If a be not greater than 2, the series (l) and (2) are 
immediately convergent. 

_ . . tanO 

lo arrive at these series we have supposed to become 

unity when vanishes, which is the case only when the angle is 

referred to the circular measure — - — . Hence a, which is 

radius 

equal to n9, must also be referred to the same measure in the 
series (l) and (2). 

102. The sine of a very small angle is equal to the angle 
itself very nearly, — the angle being expressed by the measure 

arc 
radius 

Let a be a very small angle. 

Then, by the last article 



a 3 a 5 



sin a = a - 



1.2.3 1 .2.3.4.5 



v 1.2.3 } 



88 

Now a being a very small angle, a 2 and all higher powers 
of a may be neglected when compared with unity ; 

.-. sin a = a nearly, when a is very small. 

Cor. If a and /3 be two very small angles, we have 

sin a a number of seconds in z a 
sin /3 jS number of seconds in Z j3 

103. If a be an angle so small that a 2 and higher powers 
of a may be neglected when compared with unity, equation (l) 
of Art. 101, becomes 

cos a = 1 . 

If a 2 , a 3 be retained, but higher powers of a be neglected, 
we have from (l), 

sin a = a 



1.2.3 
And from (2), 

a z 

cos a = 1 -: 

1.2 

approximations which are often practically useful. 

104. Ife*^- 1 and e~ 9xrr ^ be expanded in terms of 
y/ — l and — v — 1 m ^e same manner as e x is expanded 
in terms of x in the series, 

e * = i + a? + + — — + ... App. t. 19. Cor. 

1.2 1.2.3 rr 



then, (l) cos 6 = ± . (e 9V ~^ + e 

l 



ev 



(2) sinfl=— - ^.(yv^T- 
2V -1 V 

x 6 aevrr -i 
(S) tan = -— = . 



89 



For, e»^I- i +eV~ - — x/~ 



1.2 1.2.3 



1 + 



0* 



1.2 1.2.3 



1.2.3.4 
1.2.3.4 



+ .- 



6 ev - 1 + 6- 0VTT =2(l-~ + — - ...) = 2cos0; Art. 101. 

1.2 1.2.3.4 J 



ande e ^- e -^-i = 2 v / -l (^ - 7— - + ...) = 2\/~ r i sin0; 



1.2.3 



.-. cos^ = i(. e ; 0V -i + €- ev ^) 



■w, 



Andsin0 = — —^ ( e eV"=7 _ e -0V- 1) ...( 2 .) ; 
2 v — 1 



Also tan 9 = 



sin 1 P 0V~^I 



cos0 V^^l * 6 0VrrT + e^^ Tr i ' 
and multiplying numerator and denominator by e ev77T ? 



tan = — - 

V-i' 



1 e^vrr.! 



+ 1 



(3.) 



105. To ^roue Gregorie's series, which is 

3 5 

e evTrT + 6- 0v ~ = 2cos0 ? 
and £ esrrT -6- 0V ^ T = 2 V / ~sin^; 

.-. e eyrri = cosO + \/'^T sin 0^ 

M 



90 

.-. e e V=T - cos 0(1 + \T^\ tan 0) ; 
.-. Z £ e 0x ^ = / £ cos0 + Z £ (l +V rr ltan0) 

0v-l = l E cos0+{\/-l tan0- — '- +— --...} 

(App. i. 14. iii.) 



/ s cos + v— 1 tan + 



(tan 0) 2 _ > (tan 0) s 



and equating the possible and impossible parts of this equation, 
we have 

n ; /»■ (tan0)2 Pa"fl) 4 . fl™*)' 

= l e cos + h ... 



n ( tan6> ) 3 ( tan ^) £ 
and = tan - + - 



If tan be not greater than 1, or be any angle not greater 
than half a right angle, this series is convergent. 

Cor. Tangent of 1 a right angle, ( or tan -j , = 1 ; 



7T 11111 

-=!-- + + + ... 

4 3 5 7 9 11 



2 2 2 

= — + — + + ...; 

1.3 5.7 9-11 

= 2 ( + + + ...) , 

\1.3 5.7 9.H / 

a series which is not convergent enough to enable us to calculate 
easily the value of it. 



91 

106. To prove Eider's series for determining the value 
of 7r, which is 

7T _ /l _L_ Jt_ \ 

4 ~ \2 " 3.2 3 + 5T? 5 " *" / 

+ \s ~ 37F + STi" 3 -•■•"]■ , 

By Art. 57, 

tan" 1 1 - tan" 1 J = tan -1 ?■ = tan" 1 !, 

2 1 +i d 



1 -_JL 

.-. tan -1 1 = tan -1 1 + tan"" 1 4, 



2 



1 1 1 

and tan -1 -J- = + - ... Art. 105. 

2 2 3.2 3 5.2 5 



tan _1 i = + 

3 3 3.3 s 5.3 5 



,-. tan 1 1, 01 



,7T /l 1 1 \ 

4 ~U 3.2 3 + 5.2 5 ■" J 

+ U ~ iTi 3 + 5T3" 5 " ■" J 



a series to determine it which converges much more rapidly 
than that deduced in the corollary of the last article. 



107. To prove Machines series for determining w, which is 

- - (I _L J_ ^ 

4 \5 ~ 3.5 3 + 5.5 5 ~ "" / 

/I 1 1 \ 

V239 3.(239) 3 + 5.(239) 5 " " ' / ' 



92 



l - 



By Art. 57, tan 1 1 - tan" 1 - = tan 



1 

5 2 

— = tan -1 - 
1 3 



1 + 5 
2 1 

2 1 3 5 7 
tan -1 tan" 1 - = tan -1 = tan" 1 — 

3 5 2 17 

1 + 



3.5^ 
7 1 



tan" 1 — - tan 
17 



tan 



17 5 



9 



1 + 



- tan" 1 

7 4o 



tan l — : - tan" 1 - = tan" 1 
46 5 



5.17 
_9._ 1 
46 5 



1 + 



9 
5.46 



= tan 



239/ 



adding these equations and omitting the terms common to 
each side of the resulting equation, we have 

tan" 1 1-4 tan" 1 - = tan -1 ( J ; 

5 V2397 

.*. tan" 1 1 = 4 tan" 1 - + tan" 1 ! 1; 

5 \239J 

7T _ /l 1 1 \ 

4 ~ \5 3~7 S 5T5 7 '" / 



-[± L_ 

I 239 3.(239) 3 



+ 



.(239) 3 5.(239) £ 
In this way it is found that 

tt = 3. 141592653589793 ... 



•I 



108. The tangent of a very small angle is equal to 
the angle itself very nearly; the angle being expressed by 



the measure 



arc 
radius 



93 



^ „ „ (tan0) 3 

For = tan0 - — + ... Art. 105. 

3 

rtf (tan0) 2 > 

= tan0{l -- '- + ...} . 

* 3 5 

Now when is very small tan is very small, and therefore 
(tan 0) 2 and all higher powers of tan may be neglected when 
compared with unity ; 

.-. = tan nearly, when is very small. 

Cor. Hence, if a and /3 be two very small angles, we 
have 

tan a a number of seconds in Z a 
tan /3 /3 number of seconds in Z /3 



109. Having given tana, tan/3, tan X, to find 

tan (a + /3 + ... + X). • 

(Cos a + V - 1 sin a) (cos /3 + v - 1 sin /3) 
= (cos a cos /3-sin a sin /3) + v - 1 (cos a sin/3 + cos /3 sin a) 

= cos (a + /3) + <\/^l sin (a + /3) (l). 

Suppose this law to hold for n factors, so that 
(cosa + v -1 sina)(cos/3+ a/~ 1 sin/3)...(cos/c + V-l sin/e) 

= cos (a + /3 + ... + k) + a/- 1 sin (a + /3 -f ... + k) ; 
and introducing another factor, we have 
(cosa + \/- 1 sin a) . . . (cos k + y/-l sin k)(cosX + v-l sin X) 

= {cos(a+/3+.. 8 + K:)-|-'\/- ls i n («+i3+...f)}( cos ^+V -isinX) 
= cos (a + )3 + ... +X) +a/-1 sin(a + /3 + ... + X), by (l) ; 
and therefore the law holds for n + 1 factors, 



94 

But by actual multiplication the law has been shewn to 
hold for two factors, it must therefore hold for three factors, 
and thus by successive inductions we conclude that it holds 
generally. 

Let S 1 = the sum of the quantities tan a, tan /3, ... 

8 2 = the sum of the products of every two of them, 

#3 = the sum of the products of every three of them; 
and so on. 

Then cos (a + /3 + ... 4- X) + y/ -I sin (a + j3 +...+ X) 

= (cosa + \/-l sina)(cos/3+v-l sin /3) . . . (cos X + v-1 sinX) 

= cosa(l+\/-ltana)cos/3(l+v — ltan/5)...cosX(l-j-v -ItanX) 

= cos a cos/3... cos X(l + v-1 S 1 -S 8 - \/- 1 S 3 + S±+ ...) . 

Wood's Algebra, Art. 271. 

Hence, equating the possible and impossible parts, we have 

cos(a-f/3 +...+X)=cosacos/3...cosX(l - S 2 + S i -Se+.-.) 

sin (a+fi +...+X)=cosacos/3...cosX (S l - S 3 + S 5 - ...); 

sin(a + /3+...) Si-Ss+S*,-... 



,\ tan (a + /3 + ... + X) = 



cos(a + /3+...) 1-S 2 +S 4 - 



If there be n of the angles a, /3, ...X, it may easily be 
shewn, as in Art. 99. Cor. that 

* ( a ^ ^1-^3+^5-. .. + (-i)^~ 1 ^_ 1 

tan (a+p+...+X) = — - — '— , n even 

1 -S 2 +S i -;..+X-i)\d n 

= ^=3 9 »<>dd. 



95 



110. If sin p = sin P . sin (z + p), required to expand p in 
terms of sin P and of the sines of z and its multiples. 

Hymers' 1 Astronomy ; Art. 233. 



Since sin p = 



2\/- 1 



(eW.-a _ € --W-i) Art. 104. 



2 



the equation becomes by substituting such values of sin p and 
sin 0+i>), 



and multiplying both sides of the equation by 2 y/ _ i e pVZ "i, 
we have 

€ 2pVTT _ x _ s i n p. ( e (=+^)V31 _ e -W~^ 

= sin P. e 2VrT • e 2pV ~ - sin P . e- sV ~; 
, e *W=i(l -sinP.e sV ~) = 1 -sinP. e - sVz ^; 
._ 1 -sin P. 6- zVri 

.• € 2pV-l = _. 

1 -sin P. e 2 ^ 1 
.-. S^v^l = h (1 - sin P . e- sVZ ^) - Z £ (i - sin P . e 2 ^) 

= -sinP.€- zV ~-l(sinP) 2 .e- 22V::T -l(sinP) 3 .e- 32VrT -... 

+ sinP.€ zV ~+l(sinP) 2 .e 2zV ~ + l(sinP) 3 .€ 32V:::i +... 

Appendix n. 14. iii. 



.*. p = sin P. 



:</=! 



( e W-i_ 6 - s V-i) 



+ l(sinP) 2 7= ( € 2W-i_ 6 - 22 V3l ) + ^ > 

2\/- 1 

sin P . sin % + 1 (sin P) 2 . sin 2# + 1 (sin Pf . sin 3 s- + . 



96 



111. Having given tan r = n . tan 1, to expand Y in a 
series of the sines of the multiples of 1. 

Hymers'' Astronomy, Art. 130. 
Since tan = . _„ , Art. 104. 



l e 2d ^ + i 



e 2j'V-l_j _2Z V3l j[ 



•• — 7-7= — n. = , 

.-. 6 2Z ' V:rI {l+^ + (l-^).6 2zVrT } = l-^+(l+w).€ 2 ' V:ri 5 



1-n 



+ n 



and dividing both sides of the equation by 1 + n ; if m = - 
we have 

e 2*'V— (l + m . e 2'V— ) = m + € 2lV31 i = 6 2 Z VZT (l +m . 6 -2/V— ^ 



._ 6 2?'V_i _ 6 2^V-1 



l + ^-e 



2ZV-1 



1 +W.£ 



2*V_1 ' 






-m.€ 2zV_1 +^m 2 .e 4iV - ] !.*»».**'' 



Appendix i. 14. iii. 



,\ t -l-m 



(6 



2*V^~I _ „~-2jV-l 



+J>m 2 . 



( e ^V_i_ 6 -4?vCIj_ 



/' = 2 - m . sin 2 / + \m 2 . sin 4 / - \m\ sin 6/ + .., 



97 

112. Required the number of seco?ids contained in the 

arc 
angle which is expressed by the measure — - — . 
b r * * radius 

Let a be the required number of seconds. 

Now in the same circle 

arc subtending a" a 
arc subtending l" 1 

arc subtending a' 
radius 



arc subtending l" ' 
radius 

. arc subtending a" 

But by hypothesis, — = 0, 

radius 

and since l" is a very small angle, 

arc subtending l" 



radius 





= sin l" nearly. Art. 102; 



•. a = 



Ex. 1. Thus the number of seconds contained in the angle 

p> which, Art. 110, is expressed by the measure — - — , is 

radius 

equal to 

p sinP . t (sinP) 2 . , (sinP) 3 . 

- — 7 = - 7 .$m% + ±> — : -.sm2^ + i £- .sin 3^ +... 

sin 1 sin 1 ^ sml * sin 1 

Ex. 2. In Art. Ill we have 
The number of seconds in I' 
I' I m 



sin 1 sin 1 



. sin 2 / + 1 . — - . sin 4/ 



The angle I, which is the observed latitude of a place, is 
read off in degrees, minutes, and seconds from the instrument 

N 



98 

by which the observation is made ; therefore to find the degrees, 
minutes, and seconds in l\ we have only to determine the 
number of seconds in the latter part of the series, viz. 

m m 2 . 

; 77. sm 2l + £ — ^.sin M — ... 

sin l" 2 sm 1 



113. To expand in a series of the cosines of 

r 1 - e cos 9 J J 

9 and its multiples ; e being less than unity. 



By hypothesis e is less than unity ; and since (l — 6) 2 , or 
1 — 2 b + b 2 , is necessarily positive, 1 + b 2 is greater than 2 b. 
Let therefore 

2b 
e = 



l +¥ 
(l - b 2 ^ 



46' /1-6V 
• 1 — e 2 = 1 is 1 1 



= V^l - e 2 ; 

1 _ a/i~I 
Whence fe 2 = 



" 1+6 2 
1 - %/i - e 2 1 - Vl -e 2 1 + \A ~e 2 



1 + V^l ~e 2 1 + \A - e 2 ' 1 + \/l -V 



1 + Vi - <?' 

1 

Let 2 cos 9 — 00 + - ; 




.-. 1 — e cos 



= 1-; -. U + -), 

= —i—.{l~& (* + ") +b °~ 
1 + b" L \ x) 



1 + &* v ' V « 



l — e cos 



= (l + b 2 ) 



0+& 2 ). 



99 
l i 



1 - boo b 

] -- 

00 



(l + boo + b 2 a) 2 + o 3 # 3 -f „..) 

, b b 2 b 3 
*( X + Z + 3 + 3+-) 



(l+ 2 ).{l + 6* + &*+... 

+ (* + * + * + ...). (*> + ^) 



+ (b n + fc n+2 + 6 W+4 + ...)• f^ n + ^) 
} 



Now 



1 -b 2 

b 
1-b 2 



+ 

i + 6 2 +6 4 + ... = 

fc + & 3 + 6 5 + ... = 

l—o 

1 - = (l+6 2 ). (— L + — _.2cos0+ -.2cos20+...) 

l-ecos0 v 7 \l-& 2 l-o 2 1-6 2 / 

l+o 2 



1-6 2 
l 



VT^ 



. (1 +26. cos + 2b 2 . cos 26+...) 
=-v(l + 2o.cos0 + 2o 2 „cos20+...) 



100 





Since e is less than unity, h or . is small, and this 

series converges rapidly. 

(1 - e 2 ) 

Cor. The equation to an ellipse is r = a . ^-- , the 

^ r 1 - e cos 

focus being the pole and being measured from that vertex 

which is the nearer to the focus, 

.-. r= a . \/l - e 2 . (l + 2b. cos + 2b 2 . cos 20 + ...), 



where b = 



1 + 



x/T^T 



114. To expand 1 6 (l — e cos 0) in a series of the cosines 
of and its multiples. 

As in the last Article, 



I -ecos0 = -(l -boD).(l - -) . 

1 + b 2 ' V W 



e 1 

Where b = -, , and 2 cos — x + - ; 

1+Vl-e 2 ^ 



■.Z 6 (l-ecos0)=Z 6 r; + ^(l -ba?) + l e (l -- J 

l+o \ so) 



l 

i To 2 



Z e — jz ~6^-1.6V-J.&V-... 



6 1 ^! x h l_ 

,v * ' x 2 3 ' w :i 



^l e -o.2cos0-io 2 .2eos20 i& 3 .2cos30-, 



101 



e 
And since b — 



i+a/i-6 2 ' 



. i _ (i-vgy = cuv^y = j (1 vr=?) . 

1+6 2 (i+yx.^2^^ 2+2V ^z^ 2K 

115. To expand (a 2 - 2ab cos + b 2 ) n in a series of the 
cosines of and its multiples. 

Le^ a be greater than 6. 

Now (a 2 - 2ab cos + 6 2 ) n = a 2n ( 1 - 2 . - cos + -}. 

V a a 2 / 



Let 2 cos = ,i? + - , and - = c ; 

<2? a 



6 . o 2 \" 



1 -c 



( b « b \ 

.'. 1 -2.- cos0 + - 

V a a 2 ) 

(i- c ,)»(i_£)" 



2 2 3 7 



,2 *, i ^ o «3 



, c n - 1 c 4 » — 1 w — 2 c* 

0? 2 a* 2 3 a* ' 



Performing the multiplication, we have for the first term, 
which does not involve <#, 

/y) "I m _ 1 « _ 9 

1 + (ncf + (n . —— . c 2 ) 2 + (» ■ . . cj + ... 

7 2 ' v 2 3 



102 



The coefficient of #, and also that of - , is 

w 



n -X n - 1 n - 2 n - 1 
. c.nc — n . . cr. n. 

2 2 3 2 



— nc — n. .c.nc — n. . <r. n . c l - 



n — 1 „ / n — 1\ 2 n — 2 



or -|«c + w s . e 3 + w . . c 5 + 

* 2 V 2 J 3 



The coefficient of ,r\ and also that of — , is 



n-1 Ch n 7i-l n-2 A ( n-l\ 2 n-2 n-3 



(••=?■ 



2 3 \ 2 J 3 

and so on. 

Now ,3? + - = 2 cos 0, a? 2 + — = 2 cos 20, &c. ; 

wherefore we have 

(a 2 -2ab cos + b 2 ) n 

= a 2w (] -cat)" (l ~ -) 

f / 6\ 2 / rc-l& 2 \ 2 / n-1 n-2b" 



+ . 



2 3 a° 

6 rc-1 6 3 / »-l\« n-2tf . 

„ c 71 - 1 6 2 a w - 1 rc - 2 fe 4 

+ 2 cos 2d \n. + n*.— — .— i + -*) 

1 2 a 2 2 3 a 4 



-....) 



In a manner similar to that employed in Art. 114. we may 
expand I, (a 2 - 2 a b cos + & s ) n - 






103 



116. To find the general term of the series investigated 
in the last Article. 



This is the term involving 2 cos r6 ; 

Now f 1 - 2 - cos + —J 
V a a*} 

fl ^- X 22 n-l n -2 
= -J 1 — »Ccj? -f w . ex — n . . (far + ... 

2 2 3 

v n. (n - 1) ... (n - r + l) 

... + (- iy — i 1 — v ^—±<?a? + ... 

V ' 1.2.3 r 

n. (n - 1) ... kn - (p + r) + l} . 

' 1.2.3 (p + r) *** y 

, c n—lc 2 . ^„n.(n-l\... (n-p + l) c p . 

1 a? 2 a? 8 v J 1.2.3 p a* S 

The coefficient of <.T r , and also that of — , is 

x r 

tXr f w .(*-!)., .(w-r+1) r ?z 2 .(7z-l).(rc-2)...Q-r) . 
1 lis • C -j- x .. c + . . . 

. \ 1.2.3 r 1.2 (r + 1) 

{n-l)...(n-p + l) \ 2 (n-p).,.{n-(p+r) + l} ^ + 1 

1.2.3...p / (p+1) (p+r) J' 



n 

+ ' 



Hence the general term for the expansion of Art. 115. is 

n . (n - 1) ... (n - p + 1)\ 2 



a 2n ( _ lV ( "•("-U -.{n-p + lK 
K ' \ 1.2.3 ...p 

(n-p)... {n - (p + r) + l} /b^ 2 P+ 



(P+1) (*> + »*) \« 



2 cos r ; 



from which, by giving r the values 0, 1, 2 ... successively, 
and for each of these assumptions giving p the values 1, 2, 3 ... 
successively, we should obtain all the terms of the expansion. 
If n be an integer, the magnitudes of p and r will be limited 
by the condition that p + r is not greater than n. 



104 



117- Ex. Required to expand — — r in 

^ r (a~-2ab cos# + & 2 )* 

a series of the cosines of and its multiples. 
1 



(a 2 - 2ab cos +6 2 )* 
1 1 



a f b b*y 

1 - 2 - cos + — 
\ « or/ 

- . (1 <2?)~2 ( 1 - - . - ) , if 2 cos = a? + - 

a a \ a w) oo 



= - (1 + --30 + — -.ct? 2 +— '— — .# 3 + ...) 

a 2a 2.4 a 2 2.4.6 «° 

1 6 1 1 . 3 6 2 1 1.3.5J 3 1 

v 2 a a? 2 . 4 a 2 # 2 2 . 4 . 6 a 3 # 3 

1 f (1 by (1.3 b 2 y (1.3.5 b 3 \ 2 

\ W l_2 a \2/ 4 a° \2 . 4/ 6 a 5 J 

/ 2 1\ JT1.3 b 2 (1\ 2 3.5 b* (l.sy 5.7 b* -] 

1 r /1 6\ 2 /i.s &v 

„ rl 6 /1\ 2 3 6 3 /1.3\ 2 5 6 5 1 

+ 2cos0--+-.-- + — - .-_ + ... 
L2 a \2J 4 a 3 \2 . 4/ 6 a 5 J 

. ri.s b 2 (1\ 2 3.5 V -1 

4- 2 cos 20 — — + - - + .. . 

L2.4 a 2 \2J 4.6 a 4 J 



+ .... 



The same result might have been obtained by substituting - ^ 
for n in the series of Art. Il6\ 



105 

It is to be observed that these series are of no practical 
use unless b be much less than a. If this be the case the 
coefficients of 2 cos 0, 2 cos 20 ... rapidly decrease, and a few 
terms of the expansion may be taken for an approximate value 
of the expression which it was required to develope. 

118. If sin (a) - y) = sin &> cos u, where y is very small, 
required the value of y. 

Sin w cos y — cos &> sin y = sin w cos u ; 
dividing by cos co, and putting for cos y and cos u their values, 

1 -2 I sin - J , and 1 - 2 1 sin -J , 
tan &) . 1 1 - 2 I sin - J > - sin y = tan w . { J - 2 I sin - J > ; 

r V 2/ -> / . w\* 

.'. sin y . \1 +2 — : . tan a> [ = 2 tan w . sin - | . 

1 sin y 3 \ 2/ 

2 



2 H) 2 H) 



Now : = a= tan ^ ; 

sin«/ . y y 2 

y 2 sin - . cos - 

2 2 

w , y 

And since «/ is very small, writing y and - for sin y and tan - , 

At At 

Arts. 102. 108. 

2 tan a) 



y = 



H) 



v 

1 + - . tan w 
2 



sin - 1 . (1 — . tan ft) +.,.). 



O 



106 

Now neglecting the second and all the succeeding terms of 
the expansion, as being small when compared with unity, we 
have for a first approximation, 



y = 2 tan w 



•( sin i) S; 



And in the second member of the equation putting for y its 
first approximate value, we have for a second approximation, 

( • n \ 2 c / • ™\ 2 

y = 2 tan w . ( sin - ) \ 1 - tan o> . ( sin - 1 . tan ay] 

= 2 tan o) .( sin — J \ 1 - (tan o>) 2 . ( sin - J > . 

The number of seconds contained in y, is Art. 112. 

y f tan w ( . u\ 2 (tan w) 3 / . u\ 4 ) 

7, = % { ~ — 77 • sin - - ^ — ,,- sin - } ; 

sin l" {sinl" V 2/ sin l" V 2/ J 

and the two terms of this expression having been separately 
determined by means of logarithms, the number of seconds in 
y is known. 

Hymers* Astronomy, Art. 148. 

119. If cos (# + y) - sin n . sin % . cos m] + cos % . cos w, 
where y <md n are very small, required an approximate value 

Cos % . cos y - sin * . sin # = sin n . sin # . cos m + cos # .cos n, 
dividing by sin *, and for cos y and cos n writing 

1-2 (sin|) , and 1-2 [sin -V, 
cot*.|l-2 (sin |J |-siny=sin^.cosm+cot^.|l-2(sin-)1 ; 



107 

/ y\ 2 . ( . n\ 2 

:. 2 cot % . 1 sin - ) + sini/ = - sin n . cos w + 2 cot #. I sin - J , 



. r ( sin i)i /'■■ *y 

\ %mv\ 1 + 2 cot # . =-sinw. cos m + 2 cot * . sin — . 

*\ smt/ J V 2/ 

2 (sin - } 

V 2/ ?/ 

Now, — ■-: = tan - ; and writing the angles themselves 

sin y 2 * * 

for the sines and tangents of y and - , n and — , we have 

- n . cos m + — . cot % 

2 

2/ = 

V 
1 + - . cot % 

2 

fi 2 y 
= (- Ti.cosm h . cot#) (l .cot % + ...) 

2 7 v 2 

Neglecting — cot ss with respect to n cos m, and also - eot %+ ... 

with respect to 1, we have for a first approximation, 

y — — n. cos in. 

And in the second member of the equation putting for y 

its first approximate value, we have for a second approximation, 

. n 2 x . n 

y - (—n. cos m -\ . cot ») (1 + — . cos m . cot ss) 

2, £ 

n~ x9 nr 
— — n . cos m . (cos m Y cot % H . cot % 9 

2 \ 2 

neglecting the term involving w 3 ; 



W 



= — n . cos m h ■ . cot # . (sin m) 2 . 

2 

The number of seconds in y is 

y n.cosm n 2 .. „ 

H : — 77 . cot # . (sin my. 



sin 1 sin 1 2 . sin 1 

the terms of this expression are, as in Art. 118, determined 
separately by means of logarithms. 

Hymers* Astronomy, Art. 124. 



108 

120. To expand independently of Demoivre^s Theorem 
sin and cos in series ascending by powers of ; assuming 

,i . . , , sin0 , . 

the principle that — — = l when = 0. 
u 



Let sin = A + a x + a 2 a + fyO 3 + ... where J x a 19 a 2 , 
are constant quantities whose values are to be determined. 

Now sin = - sin (- 0) ; 

.-. A + «i0 + a 2 2 + « 3 3 +'—'=- (^ - «i0 + «2# 2 - «3^ + .. 

whence -4 + a 2 6 2 + a 4 4 + ... = 0, 

and if = in this equation, A = 0; 

,\ o 8 8 + a 4 4 +... =0; 

.-. a 2 + a±0 2 + ... =0, 

And if = ; a 2 = 0. 

In like manner a 4 , a 6 , a 8 ,... may each be shewn to be = 0. 
sin0 



• • • 



e 



«i + « 3 3 + 



_ a ._ _ sin0 

But if = 0, -— = l. Art. 100. .-. 1 = a v 


.-. sin0 = + ag^ 3 + a 5 5 + ... 

Again, let cos = J5 + b x + 6 2 2 + ... 

And if = 0, cos = l, .*. l = B, 

Also, cos = cos (— 0) ; 

.-. 1 + M + 6 2 2 + ... m 1 - M + b 2 0* - M 3 + ... 

And as before it may be shewn that b x , 6 3 , 6 5 . . . are each = ; 

.-. cos0 = 1 +b 2 2 + biO i + ... 



109 

Adding these values of sin and cos 0, we have 

l+0 + & 2 2 + a 3 3 + & 4 4 + ... = cos0 + sin0; 
writing + <j) for 0, 

i + (0 + 0) + h (0 + 0) 2 + 3 (0 + <t>y + ... 

= cos (0 + 0) + sin (0 + 0) 
= cos (cos + sin 0) + sin <p (cos - sin 0) 
= (1 + htf + ...) (1 + + M 2 + « 3 3 + ...) 
+ (0 + « 3 3 +...) (1 - + 6 2 2 - « 3 3 + ...), 
and equating the coefficients of <f> 0, d)0 2 , cf)0 3 , 00 4 , we have 
2& 2 = - 1; .-. b 2 = - 



3a 3 = 6 2 ; .-. a 3 = - 

4& 4 = - a 3 ; .*. 6 4 = + 
5a 5 = 64 ; ,\ a 5 = + 



1.2 

1 
1.2.3 

1 
1.2.3. 4 

1 

1.2.3.4.5' 



and so on ; — the law of the coefficients being manifest. 

Hence sin = - + ... (1.) 

1.2.3 1.2.3.4.5 v ' 

Q2 A4 

and cos 0=1 + ... (2.) 

1.2 1.2.3.4 v ' 



CHAPTER VII. 



ON THE SOLUTION OF EQUATIONS AND THE RESOLUTION OF 
CERTAIN EXPRESSIONS INTO FACTORS. 



121. To solve a quadratic equation. 
i. Let w* + *poc - q = be the equation. 



»-.--&*/■&$■- j'&^fr')} 





Let if- 

2> 


: (tan 0) 2 






(0 


,*. OG - 


-v^.| 


[tan 


%/{0 


tan 0) 2 + 
tan 


^} 


= 


- vr?.- 


=f sec0 
tan0 










/- COS =j= 1 








cos 0—1 


•(- 


. 0\ 2 
sin - 

2/ 



= — tan ~ , 

2 




sow . 

sm0 


2 sin 



- . cos - 

2 2 




cos 0+1 


2 1 cos 

. 

2 sin - . 

2 


0\ 8 

cos- 

2 


1 

tan - 

2 






sin 





Ill 



If therefore ■!v l and x 2 be the two roots of the equation, 



at, = v</ tan - 
i v i 2 

x 2 = - y/q cot 



= ~\Tq 

2 

tan- 

2 



(2). 
(3). 



From (1) we have, L tan 9 = l w 2 + \l w q - l w p + 10, 



(2) 



(3) 



which determines ; 

e 



ho®i = ihoq + £ tan 10, 

2 

a 
ho( - O = i ^io^ - Ir tan - + 10. 



ii. If the equation be oo 2 — poo — q = 0, the roots are the 
same as those of the preceding equation with their signs 
changed ; for the product of the roots with their signs changed 
(- q) remains the same, and the sum of their roots with their 
signs changed (the coefficient of the second term) changes its 
sign, but continues of the same magnitude. (Wood's Algebra, 
Art. 271). 

in. Let the equation be w 2 - pa? + q = 0, 



»!{-v/(.-$}. 



4>q 4,0 

If — be less than unity, assume -~ = (sin Of 
p i J p 2 v y 



0). 



Then, oc l and w 2 being the roots of the equation, 

p I 0\ 2 

a? l = -. (l + cos0) =p. ( cos -I . (2). 

p ( 0\ 2 

a? 2 = -• (1 - cos $) = p. f sin -J . (3). 

From (1), L sinO = l l0 2 + 1 l lo q - l lo p + 10, 



(2) l lo a? 1 = l 10 p -\- 2 L cos 











20, 



(3) Z 10 # 3 = l lo p + 2 Z, sin 20. 

2 



112 

iv. If the equation be w 2 +pw + q - 0, the roots are the 
same as those of the last equation with their signs changed. 
(Wood's Algebra, Art. 281). 

4*0 4}Q 

In the two last cases if -~ be > 1, let -4 = (sec 0) 2 ; 

then < a?=-{l± A /-[(sec^) 2 -l]}=?.{l±tan0v /rr l} 5 

and both roots of the equation are impossible. 

These solutions may be employed in preference to the 
common method of solution when p and q contain a great 
number of figures. 

Ex. Required the roots of a? 2 +365*42 #-3469*1 = 0. 
By case (i) we have 

L tan 9 = Z 10 2 + 1 . l l0 34>69'l - ho 365*42 + 10. 
By the Tables, 

Z 10 2 = -3010300 
±l lQ 34>69'l = \ . (3-5402168) = 1*7701084 

10- 



12*0711384 
Subtract l 1Q 365 *42 = 2*5627923 



9*5083461 =Z tan 17°, 52' nearly ; 
'• h^i = | • *io 3469*1 + £ tan 8°, 56' - 10. 
l.Z 10 3469*l = 1-7701084 
L tan 8°, 56' = 9*1964302 



10*9665386 
Subtract 10* 



•9665386 = / 10 9*2584 nearly. 



113 



Again, l l0 (-a? 8 ) = 1 . l lQ 3469*1 - L tan 8°, 56' + 10. 
10 + \ . Z 10 3469*l = 117701084 
X tan 8°, 5& = 9*1964302 



2-5736782 = Z 10 374*69 nearly. 

Hence the approximate values of the roots of the proposed 
equation are 

9*2584 and - 374*69- 

We might, after finding a? r , have determined w 2 from the 
equation 

- oc x - w 2 = 365'4<2* 

Wood's Algebra, Art. 271. 

122. To solve a cubic equation. 

Let the equation when transformed, if necessary, to an- 
other which wants the second term (Wood's Algebra, Art. 284), 
be 

a? — qx — r = 0; 

if a? = - , it becomes 
n 

y z - qn 2 y — rn 3 = 0. 

lT _ cos 3d) 

Now (cos (p)* - J cos <p -2- = 0. Art 48. 

And these equations are identical if 
(1) cos (p = y. 



(2) § = <^ 2 ; 


q 


(3) ® = rn 3 ; 

4 


r / 27 

,-. cos 3d) = 4rw 3 = - 'v — 

T 2 q 3 



Let « be the least value of 3<p which satisfies the equation 
(3) ; then one value of y is cos <£, or cos - . 



114. 



Also since, Art. 91, cos a - cos (2 W7r±a), the two other values 

. , , , „ 2m7r±a 
ot y are contained among the values of cos . 

o 

Now m being an integer must be of one of the forms 3p, 
3p±l; 

2.3p.7T±a ( a\ a 

and cos = cos ( 2pir ± - 1 = cos - . Art. 91. 

3 \ 3/ 3 

2(3»±l)7r±a / 27r±a\ 2?r±a 
COS = COS I 2p 7T ± — I = cos . 

3 \ 3)3 



Wherefore the three values of y are 

a 2-7T + a %ir — a 
cos - ; cos ; cos ; 

3 3 3 

y 

and the three values of #, or - , are 

n 

/q a A / q 2?r+a /q 2tt - a 

2\/-.cos-; 2 \/ - . cos ; 2 V z -cos . 

v 3 3 3 3 3 3 

123. 77ie solution of a cubic equation given in the last 
Article applies to those cases only where all the roots are pos- 
sible: i. e. to the irreducible case of Cardan's rule. 

Wood's Algebra, Art. 331. 



Since cos a is always less than 1 



q 

r 

- < 
2 

r 2 a 3 

- < — 



V q - 



4 27 
which is the irreducible case of Cardan's rule 



115 



124. To resolve the equation x 2n - 1 = 0, into its qua- 
dratic factors. 

x 2n -1 = 0; 
.\ «2? 2w = 1. 

Now (cos0 + V- 1 sin0) 2w = cos 2n0 + \/- 1 sin 2n6, 



mir 



and if — , m and n being integers, 



n 



KYlTT / . m<7T 

cos — + v - 1 sin 

n n 



cos2m7r + y/ — 1 sin2m7r 



= 1 



mir 



lib 7T J 

.*. o? = cos + v 



1 sin 



. m7r 



w 



Now m being an integer must be of the form p . 2 n + r, 
where p is ? or any integer, 

and r is 0, or any integer less than 2n. 



TT (p • 2n + r) 7T j . (p.2n + r)ir 

Hence oo — cos — h v — 1 sin — — 



= COS(2J0 7T + — J + V - 1 sin \2p7r + — I 



rir 



cos 



— + V - 1 sin — . Art, 91 



Tic 

n 



116 

And for r writing 0, 1, 2, ...2n— 1, successively 

(1) If r = 0, a? = 1, 

(2) r = 1 , a? = cos — f- \/ - 1 sin - , 



(3) r = 2, a? = cos — + v / -lsin — 



(n + 1) r = w, a? = - 1, 



%fl 2 / 2 7Z — 2 

(2n - l) r = 2n — 2, w = cos 7r + v — 1 sin it 

n n 



2tt / . 2tt 

= cos v — l sin — . Art. 91. 

n n 



(2n) r = 2w - 1, a? = cos w + y/ — 1 sin — tt 

n n 



IT r . 7T 

= cos v - 1 sin — . 

n n 

Now the equations (l) and (n + l) give the quadratic factor 

(a? - 1) . (a? + 1) = a? 2 - 1 ; 

the equations (2) and (2n) give 

/ 7T V . TT\ ( IT J . 7T\ 

a? - cos — - v - 1 sin — ) a? — cos \- \/ - 1 sin — 

V W n) \ n n) 

= a? 2 — 2 ( cos — J . ,i? + 1 , 
V n) 

and so on ; 
wherefore we have 

ff 8 »-l = (a? 2 -l). {a' 2 -2(cos-).a;+l}...{a ,2 -2fcos- -7r).«+l] 



117 

125. To resolve the equation x 2n + 1=0 into its qua* 
dratic factors. 

v~ n + 1 = 0; 

.-. cn 2n = - 1. 
Now (cos# + \/~^~l sin0) 2w = cos 2nO + \/ - 1 sm2n6, 

making 2nd = (2m ■ + l)7r, and proceeding as in the last Article, 
we have, 

2 m + 1 / . 2m+l 

CO = COS 7T + V — 1 Sin 7T. 

2n 2n 

Also, assuming rn = p . 2n + r, where p is 0, or any in- 
teger, and r is 0, or any integer less than 2n. we have, as 
before, 

2 r + 1 j . 2 r + 1 

CO = COS 7T + V — 1 Sin 7T- 

2n 2n 



(1 ) If r = 0, co = cos i-v-1 sin — ; 

v ' 2n 2n 

Stt / . 3tt 

(2) r =1, cc = cos — + V - 1 sin — ; 
7 2^ 2?z 



2n - 1 / . 2n - 1 

(n) r = n — 1 , a? = cos ir + v - 1 sin tt ; 

v ' 2n 2n 

2n + 1 / . 2n+\ 

(n + l) r = n, co = cos 7r + v — 1 sin — — ir 

y J 2n 2n 

2n-l / — . 2n-l 

= COS(27T 7r) + V -lSin(27T 7r) 

2n -1 y , 2n - 1 

= cos 7r - v - 1 sin ■ 7r ; 

2n 2n 

(these two are the middle terms of this series of equations) 



118 



, 4>n - 3 j . 4<n - 3 

(2n-l) r=2w-2, # = cos it + v - 1 sin tt 

2n 2n 



3tt j . 3tt 

= cos v - 1 sin — 

2n 2n 



(2n) r=2w-l, <a?=cos 



2n 



— . 7T 

1 sin — 
2n 



And forming quadratic factors out of pairs of simple 
factors equidistant from the extremities of this series of equa- 
tions, we have 



7T j- 7T \ 7 7T 

since I <3?-cos \- v — 1 sin — • \w — cos — 

2n 2n \ 2n 



1 sin 



2w 



x 2 — 2 ( cos — ) a? + 1 ; and so with the others ; 
V 2nj 



M ln + 1 = \w' d — 2 I COS — - I . a? + If . •) .# - 2 I COS J . a? + 1 



2w 



2rc 



f 9 , 9,71 - 1" 
. -; a- — 2 (cos 7t) . a? + 1 

v 2rc ' 



126. To resolve the equation x~ n + 1 — 1=0 m£o its 
quadratic factors. 

As in the two last Articles, 
(cos + y/^l sin 0) 2 * + 1 = cos (2w +1)0+ ^^ sin (2w + 1) 



= 1, if (2 w + 1) = 2nnr, or = 



2mir 

2?l+l 



,2n + \ 



2 m / . 2 m 

'. 30 = COS 7T + v - 1 sin ' 



2n + 1 



2w+l 



119 



And if m = p. (%n + l) + r; where p is or any integer, and 
r is or any integer less than 2w+l, it may be shewn as 
before that 



2rw 
X = cos f- 

2/2 + 1 



l sin 



2?-7T 
2 71 + 1 



(l) If r = 0, x=i 



CO 



r =1, x — cos 



2tt / . 2tt 

V - l sin 



2n + 1 



2n+ l 



/ . 2/27T y . 2727T 

(n +1) r — n, a? = cos \- v - 1 sin 

2/2+1 2w+ 1 



(n + 2) r=ra+l, a? = cos 



2w + 2 y 2 ?2 + 2 

7T + V - 1 Sill 7T 

2n + l 2w + l 



2/Z7T 
= COS J 2-7T i + 

2%+l 



- . / 2nir\ 
1 sin 2tt-- ! 

V 2/2+1/ 



= COS 



72+1 



-v 7 - 



l sin 



272 + 1 



(2n + l) r=2/2, x = cos 



2/2 + 1 



^ 



1 sin 



2tt 



27i + 1 



Hence one root is 1, and a factor of the equation is a? — 1. — 
and by forming quadratic factors out of equations (2) and 
(2n + 1), (3) and (2n), ... we have 



1 = (a? - i) . ja? 3 -2 cos 



2tt 



222 + 1 



. X + 1 



2?l7T , 

. . J ar - 2 J cos 1 . x + 1 £ 

V 2tt + 1 



120 

127- To resolve the equation x 2n+1 4- 1 =0 into its 
factors. 

Here w 2n + l = - 1, 

(cos + \/- 1 sin 0) 2n+x = cos (2 n + l) + a/ '- 1 sin (2rc + l) 0, 

and let (2w+ 1)0= (2m + 1) tt, or = tt ; 

2rc + 1 

... (cos + a/^ sin 0) 2 « +1 = - l 



7T + V- 



.*. oo = cos 7r + a/ — 1 sin 



9,71 + 1 2/2 + 1 

Proceeding as in the last Article by making m=p.(2w + l) + r, 
we have 

2r + 1 /- — . 2r + 1 

0? 5= COS 7T + V - 1 Sin 7T. 

2w + 1 2w + l 



7T / . 7T 

sin 



7j- > 

(1) If r = 0, o? = cos + v-lsi 

27i + 1 2^ + 1 

3?r / • 37r 

(2) r=l, o? = cos +v-lsin 

7 2^+1 2w+ 1 



(n + 1 ) r = w, a? = cos 7r -f- a/ - 1 sin 7r = - 1 , 



, v 3tt y . Sir 

(2w) r=2w-l, a? = cos \/-lsin 

2^ + 1 2n + 1 

7T y . 7T 

sin 



(2n + 1) r = 2w, o? = cos _*/_] s i 

2W+1 v 2w -f 1 

and we have as before, 

07 2w + 1 +l = {W+1}. {o7 2 -2 (COS — ) .07+1} 



2w - 1 
07 2 - 2 (cos tt) . x + 1 1 

2w + 1 ' 3 



121 
128. To resolve sin and cos into factors. 
The values of which satisfy the equation sin = are 

0, ± 7T, ±271", =J=37T, ±^7T, ... 

n being any integer whatever. 

Hence the series which expresses the value of sin0 is 
divisible by 

0, 0-7T, 0+7T, 0-27T, + 27T, 

Wood's Algebra, Art. 269. 

Let therefore 

sin0 = a.0(0-7r) (0 + tt) (0-2tt) (0 + 2tt)... 

where a is some constant quantity whose value is to be 
determined. 

.. sin0 = ±a.0(7r-0) (tt + 0) (2tt-0) (27T + 0)' 

0\ / 0\ / 0\ / 



L ±a V.2V.3V. & c. (l-5)(l-^) 



.„ „ sm 

Now if = 0, — — = 1, Art. 100. 


making therefore = 0, our equation becomes 

l = ±a.7r 2 .2 2 7r 2 .3 2 7r.&c. 

**"(-3 (-&)(■-£) » 

Q 



122 

Again, the values of which satisfy the equation 
cos = are 

ir 3tt 5-tt 2rc+ 1 

2 2 2 2 

w being any integer whatever. 

Hence we have as before 
cos « ± 6 . — . 



2 2 2 2 



/ 2 2 2 \ / 2 2 2 \ 



Now if = 0, this equation becomes 



^2 o2 ^2 
1 = ± ft . — . — - 



2* 2 
2 2 



COS 






APPENDIX I. 



ON THE LOGARITHMS OF NUMBERS, AND THE CONSTRUCTION OF 
THE LOGARITHMIC TABLES OF NUMBERS. 



1. Def. If n = a x , x is called the logarithm of the 
number n to the base a ; or, the logarithm of a number to a 
given base is that power to which the base must be raised 
to give the number. 

The logarithm of n to the base a is thus expressed, l a n. 
Wherefore if n = a*, then os = l a n, and n — a x — a hn . 

2. If, while a remains the same, successive values be 
given to n and the corresponding values of w be registered, 
the tables so formed are called " tables of a system of 
logarithms to the base a". 

Since, as will hereafter be shewn in Art. 10, a system of 
logarithms to the base 10 is attended with peculiar advantages, 
the tables of logarithms in common use are calculated to that 
base. 

3. Required from tables of logarithms calculated to a 
given base, as e, to form tables of logarithms to any other 
base, as 10. 

!n = e hn , 
n = 10* lon 
hio . 



10 = € 

pten _ jo?io« _ l£hl0\l l0 n _ ^hlO.l^n . 

.-. l £ n = l e 10 .l lQ n ; 



124 

Hence, the logarithms of any number n in two systems 
calculated to any bases, as 10 and e, are connected by a 

constant multiplier, viz. - — ; — and therefore from tables 

Z.io' 

of logarithms calculated to a base e we may form tables to 
the base 10. 

By writing a for 10 in this proof, we get l a n = — . l e n, 

l g a 

and the constant multiplier connecting the logarithms of a 

number in the two systems is — . 

l e a 

4. The logarithm of the product of any number of 
factors is equal to the sum of the logarithms of the several 
factors. 

For m.n.r ... = a l a m . a l <* n . a laT . . . 

— a (!a m + l an+la r + • • •) 

But m.n.r ... = a laimnr -" ) ; 
.-. l a (m .n .r ...) = l a m + l a n + l a r + .... 

Hence, if we find in the tables the number whose logarithm 
is the sum of the logarithms of the several factors, we obtain 
the product of those factors. 

5. The logarithm of a quotient is equal to the logarithm 
of the dividend minus the logarithm of the divisor. 

i a (5) m a l * m . . 
For a aKnJ = - = -— = *a»-i*» ; 
n a l <* n 



(m\ 



Hence, if we find in the tables the number whose logarithm 
is the logarithm of the dividend minus the logarithm of the 
divisor, we obtain the quotient. 



125 



6. The logarithm of the c th power of any number is equal 
to c times the logarithm of the number, — c being whole or 
fractional. 

For a 1 ^ = m c = (a l <* m ) c = a cA « m ; 

Also ^« (w)7 = m* = (a z « m F = ai' 1 *" ; 
i l 

C 

Hence the c th power of any given number is that number 
in the tables whose logarithm is c times the logarithm of the 
given number ; and the c th root is that number in the tables 

whose logarithm is ( - ] of the logarithm of the given number. 

7« From the last three Articles it appears that tables of 
logarithms enable us, particularly where the numbers are large, 
to perform the operations of multiplication, division, involution, 
and evolution, with greater ease than by the common arith- 
metical methods. The easier arithmetical operations of addi- 
tion and subtraction cannot be performed by logarithms. 

8. In the common, or Briggs\ system of logarithms, where 

N 
the base is 10, the logarithms of 10 n . N and — - may be deter- 
mined at once from the logarithm of N. 

For l 10 (10 M . N) = l l0 10" + Z 10 JV Art. 4. 
= n.l 10 10 + l 10 N Art. 6*. 

= n + l 10 N. 



And /, 






l 10 N- l 10 lO n Art. 5, 



ImN- n, 



126 

Thus we find in the pages of logarithms printed at the end 
of this Appendix, 

l l0 6 = 07781513; 
.-. l 10 60 = Z 10 10 + l 10 6 = 1 + 0-7781513 = 1*7781513, 
Z 10 600 = / io (10 2 ) + l 10 6 = 2 + 0-7781513 = 27781513, 
l l0 -6 = Z 10 6 - Z 10 10 = 0-7781513 - 1, 
J 10 -006 = ho® ~ ^io(10 3 ) = 0-7781513 - 3. 
The last two logarithms are thus written, 17781513, 37781513. 

Def. The decimal part of the logarithm is called the 
mantissa of the significant digit 6 ; the integral part is called 
the characteristic. 

Thus, in Z 10 600 = 2*7781513, 2 is the characteristic of the 
logarithm of the number 600, -7781513 is the mantissa of the 
significant digit 6. 

9. In the common system to determine the characteristic 
of the logarithm of any given number. 

If a number be between 

1 and 10, its log. is between and 1; .-. its characteristic is 0. 

10 and 100 1 and 2 ; .- , 1. 

lOOandlOOO, 2 and 3 ; .* 2. 

lO^-'andlO" w-landw; .- (n-l). 

Hence the characteristic of the logarithm of a number having 
n integral places is n - 1, or is less by unity than the number 
of integral places which the number contains. 



127 

Again, if the number be between 
1 and — , its log. is between and — 1 ; .*. its characteristic is 1, 

— and , -land -2; .*. 2, 



10 100 ' 



— and — 

10 n 10 



and ^ l 9 -n and -(w+l); .* (w+1), 



Hence the characteristic of the logarithm of a decimal frac- 
tion having n cyphers after the decimal point is (n + 1). 

Generally therefore, the characteristic of the logarithm of 
any number is the number of its digits minus unity, — where if 
the number be a decimal fraction, the cyphers which follow 
the decimal point are alone counted and are reckoned nega- 
tively. 

And conversely, if logarithms be given having characteristics 
1, 2, 3 ... I, 2 5 3 ... there are in the numbers to which these 
logarithms belong 2, 3, 4... 0,-1, —2... digits respectively. 

Thus the logarithms of 254 and 25400 have for charac- 
teristics 2 and 4, and the characteristics of the logarithms of 
2-54, 25*4, -000254 are 0, 1, 4. 

The mantissa given in the tables for Z 10 3652 is -5625308 
(p. 147). 

.-. l l0 3652 = 3-5625308, l 10 36.52 = 1.5625308, 

l 10 365200 = 5-5625308, l l0 '3652 = T'5625308, 

l 10 '003652 = 3-5625308, 

10. From the last article it appears that if the base of 
the system be 10, it is requisite to register the mantissce only 
in the tables, because the characteristics can be determined 
by counting the digits in the number whose logarithm is re- 
quired. This omission of the characteristics renders the com- 
mon tables less bulky than those calculated to any other base. 



128 



Also from Art. 8, it appears that in this system the man- 

N 
tissa of N is also the mantissa of 10 n . N and of — , where n is 

10"' 

any integer :~ this circumstance renders the common tables 

more comprehensive than if any other base were taken ; — for 

if any other base were used, the mantissa of N would not be 

n 

the mantissa of 10 w . N, or of — . 

10 n 

In the same way it would appear that if our arithmetic 
were duodecimal, tables calculated to the base 12 would pos- 
sess the same advantages which we have here shewn to belong 
to the common tables. 

11. The tables of logarithms in common use register either 
to Jive or to seven places of decimals the mantissae for numbers 
from 1 to 100000. There are printed at the end of this Appendix 
two pages of logarithms in which the mantissae are calculated 
to seven places of decimals. The line at the top of the second 
page begins with the number 3650, or 36500, and its mantissa 
•5622929 is placed opposite to it. And because the mantissae of 
all numbers from 36500 to 36559 (comprised in the first six 
lines of the page) have the same initial three figures, viz. -562, 
these three figures are registered once for all opposite to the 
number 3650, and the four last figures of each succeeding man- 
tissa are placed under the number to which they belong. 
Thus the first line of the page 



Num. 


1 ° 


1 


2 


3 


4 


| 5 


6 


7 


8 


9 


3650 


1 5622929 


3048 


3167 


3286 


3405 


3524 


3642 


3761 


3880 


3999 



gives us, mantissa of 36500 = *562 2929 

36501 = -562 3048 

36502 = -562 3167 

36503 = '562 3286, 

and so on. 

The next line in the same manner gives the mantissae of 
numbers from 36510 to 36519 inclusive. 



129 

12. Since a change in the value of the three initial 
figures may take place in the middle of one of the lines, in 
taking the mantissa of a number it is necessary to observe 
whether we ought to take the three initial figures which im- 
mediately precede, or those which immediately follow it. Thus 
the mantissa of 36643 is -5639910, and the last four figures of 
the mantissa of 36644 are 0029 ; the first three figures therefore 
of the mantissa of 36644, and of those numbers which im- 
mediately succeed it, are not '563 but -564, and the mantissa 
of 36644 is '5640029. Similar changes of the initial three 
figures of the mantissa occur at the numbers 36729, 36813, 
36898, 36983, and are indicated by printing in a smaller type 
the fourth figure of the mantissa? of those numbers. 

The construction and use of the small tables in the last 
column of the page will be explained hereafter, 



13. Examples. 

(l) To multiply 23 by 16. Art. 4. 
By the tables, p s 146., Mantissa of 23 is -3617278, 
Mantissa of 16 is -2041200 ; 
.-. Z 10 23 = 1-3617278 
Z 10 16 = 1-2041200 



2-5658478 



And, p. 147, the significant digits corresponding to the mantissa 
•5658478 are 3680 ; 

i'. l l0 368-0 or Z 10 368 = 2-5658478 ; 

And 368 is the product sought. 

R 



130 

(2) To divide 3672 by 51000. Art. 5. 

l w 3672 = 3-5649027 p. 147. 
l l0 51000 = 4-7075702 p. 146. 



2*8573325 



And, p. 146, we find this mantissa to be that corresponding to 
the significant digits 72 ; 

.-. 2*8573325 = l l0 -072 ; and -072 is the quotient sought. 

(3) To find the values of (15.4) 3 and (650)1 Art. 6. 

Z 10 15'4 = 1-1875207 p. 146. 
3 



3-5625621 = l w 3652-3 nearly, p. 147- 
.-. 3659,-3 is the approximate cube of 15*4. 

Again, Z 10 650 = 2-8129134, p. 146. 

1 

.\ -. Z 10 65O = -5625826 = Z 10 3'6524 nearly, p. 147- 
o 

.'. 3-6524 is the approximate fifth root of 650. 

14. To expand l a (l+x) in a series ascending by 
powers of x. 

Now a = 1, or l a l = 0; and l a (l + w) becomes l a l when 
as vanishes, — the series therefore which expresses the value of 
l a ( 1 + oo) cannot involve any negative power of <#, (or it would 
become infinite when oc vanished,) nor can it have a constant 
term, (or it would not vanish along with x). 



131 

Let .-. Ax + Bx 2 + Cx 3 + ... = l a (l + a?) (l.) 

so J# + Z?# 2 + Car 3 + ... = l a (l +%) 

.'. A(x-z)+B (x* -z z ) + C (x 3 - s 3 ) + ... 
«J.(l+»)-f B (l+«) 

1 +# 



= *« 



1 + # 



A x — % /«» — %\ a 
= A- + B[- +...by(l.) 

And dividing by x — ss 9 

A+B(x+%) + C(x*+xz+z 2 ) + ...=A +B , X ~* +.. 

l+# (l+*) 2 

and if % = a?, 

l +# 7 

Equating coefficients of like powers of x, 

2B=-A, 3C = A 9 4<D=-A, &c. 



2 3 4 



/y»* 'V**' *>i* 

,-. I.(i +*) -A (»_- + --'-+ ...) (i.) 



132 

Con. 1. Since A is a constant quantity, let e be the base 

of that system in which A = , 

l € a 

> l . of a? . ... x 

•■■ , -( 1 + # )-u- ( '"i' + 7-;^ (u) 

and if a becomes e, 

**(! + *)«*-- +— — (I".) 

We shall hereafter find the value of e, which is a con- 
stant quantity 2*718281 8... ; logarithms to the base e are called 
Napierian, from Napier, the inventor of logarithms, who 
adopted this base because logarithms to it are more easily 
calculated than those to any other base,' — as is evident from 
comparing the series (ii.) and (iii.) 



Cor. 2. l € a = / 6 {l + (a - l}, which by (iii.) is 

=:(a-l)-l(a-l)* + l(a-l) 3 - (iv.) 

Cor. 3. A = — = 



l e a (a-i)-l(a-i) 2 + i(a-l) 3 - ... 



Def. The quantity A, or — , is called the modulus of the 

t € a 

system whose base is a. 

15. To determine when the series (iii.) begins to con- 
verge. 

The magnitudes of the n th and the (n + m) th terms are 

of 1 w m+n 

— and , and if the series becomes convergent after 

n m + n 

the n th term, every term after the w th is less than it : 



133 



m + n n 

m + n m 

,.. w m < < 1 + - 

n n 



.'. 00 < I 



1 +- . 
n 



Similarly (iv.) becomes convergent after the n term if 



m 



(a - l) m < 1 + - . 



n 



16. We shall next investigate some formulas for the cal- 
culation of logarithms which converge rapidly. 

-o /-»x , x i ( *-l\ «- 1 i /a?-l\ 2 

^w-'tH'-tJ^t-Ht)-' 



But Z e - = Z 6 1 — l 6 x = — l e a? ; 

00 



07-1 /a?- iy 

Again, 4(1 + #) = « - - + - - ... 

w * 2 * 3 

eV ' 2 3 

1 -4- !?* 2j i^ 

- / e (l + *)-Z e (l-«), or Z 6 - = 2 .{# + - + ~ + ...}. 

1 — % o O 

. , .„ 1 + # , a? - 1 ... 

And if = oo and .-. % = — , this becomes 

1 — % oo + 1 

If co be a little greater than unity this series converges 
rapidly. 



134 



2 2 

in, l € x = l e (x m ) m = m . l € at m 



= m{(^-l)-l(^-i) 2 + ...}, by (iv.) 

and if a? be greater than unity, by assuming m of sufficient 

2 
magnitude a? w may be made to differ from 1 by any quantity 

however small ; in which case the succeeding terms of the 

series may be neglected with respect to the first, and we 

have 



l e x = m (x m - 1) 



(vii.) 



17. Having given l e x to find 1 6 (x + z), z being small 
when compared with x. 



h(x + %) = l e a?ll + -j 
t x+lJl + ?j 

(-3- ,/K)- 



= l e X +2; 



By (vi.) 



,( l+ 3 +i h i+ 3 +i/ 

= ^ + 2 {( 2 -^) + +(^T-J +i (i^) 5+ -} (viii ° 

Cor. If % - 1, 

^e(l+<^)=4^ + 2i +4-7 -, +4-7 - + ...l....(ix.) 

' \2a? + l 3 (2# + l) 3 °(2# + l) 5 J v ' 

which is useful in computing l € (l + <#) from / e a?, particularly 
when cv is large. 



135 



19 Having given the Napierian logarithms of two suc- 
cessive numbers x - \ and x. fa find that of the number next 
fall awing. 

,r - l 



— 'e 



X - 1 



Rxpanding / J l — >] bv ^vi.) we have, 



(■-5 



Ih , 



since — _ - - 

8* f - l ' 



{■41- 

I9« FVj expand a v in (; series ascending by powers of I 
* - 1 (,v - O ( r -8} 

= 1+ir(a _ 1) + (r __ ((I _ 1)2 + >r i__JL_ r i ( , 7 _ ir+ ... 

lot the coefficient of*, which is(a-l)^(a- 1) 1 ;V-.... 

be represented bv p n and let the coefficients of .r. x r \ ... be 
represented bv />,. p, j — then we have 



136 

a* - 1 + p x oo + p 2 x 2 + p 3 a? + ... 
.'. a z = 1 + pi% + p 2 % 2 +p 3 %' s + ... 

Now aP + z) = a* .a z , 
or 1 +^ 1 (<2? + ^) + 2> 2 (a? + #) 2 + + ;>»(# + #)" + ... 

and equating the coefficients of the terras involving ##, <# 2 #j 
#?%, ... ,2? w ~ l #, ... we have 

(Pi) 2 



3P* = P1-P29 ."• #> = 



2 

(?i) 3 
1.2.3 

fa) 4 

1.2.3.4 



fa)* 

n-Pn=Pl-Pn-l, -"- P," 



1 .2 .3...^ 



Nowp! = (a-l) -l(«-l) 2 + l(a- l) 3 - ... =^ 6 a by (iv.), 

1 1.2 1.2.3 1.2.3...7Z v ' 



Cor. If a become e, we have 



a? <a? 3 w n , „ . 

e * = 1 + a? + + +... + + (xn.) 

1.2 1.2.3 1.2...W V ' 



137 

20. To find the value of e the base of the Napierian 
system of logarithms. 

If a; = 1 and a become e, the series for a x in the last Article 
becomes 



1 1 


1 


lT 1.2 T 1.2.3' ^1.2. 3...n ' 


Now, l + l 


= 2 


1 
T~2 


= -5 


1 


- '1666666666 


1.2.3 




1 


- -0416666666 


1.2.3.4 




1 


— '0083333333 


1.2.3.4.5 




1 


= -0013888888 


1.2.3.4.5.6 


1 


= -0001984126 


1.2.3.4.5.6.7 


1 


= -0000248015 


1.2.3.4.5.6.7.8 


1 


= -0000027557 


1 .2.3 8.9 


1 


= -0000002755 


1.2. 3. 4.... 9. 10 




2-7182818 



This gives the correct value of e, the base of the Napierian 
system, so far as the figures are put down. By taking more 
terms and a greater number of figures in each, we could de- 
termine the value of e to any required degree of accuracy. 

S 



133 



21. On the construction of the common tables of loga- 
rithms. 

From one of the series (v), (vi), (vii), the Napierian loga- 
rithms of low primes may be found. The logarithm of a high 
number which is not a prime may be determined from those of 
its factors by resolving the number into powers of prime factors ; 

Thus, l e 288 = l e 2 5 . 3 2 = l 6 2 5 + l e 3 2 = 5 . l e 2 + 2 . l e 3. 

And Arts. 17, 18 give the expressions (viii), (ix), (x), 
which facilitate the finding of the logarithms of high numbers 
which are primes. 

The Napierian logarithms being determined, the tables to 
base 10 are deduced from them by multiplying each by , 

V£ 10 

which is equal to -434294819.... Art. 3. 

Some of the artifices used in computing the tables may be 
found in Sharped " Method of making Logarithms" prefixed 
to Sherwin's Tables. 

22. The common tables contain the mantissa? of the loga- 
rithms of numbers of five places of figures, these mantissa? being 
computed to seven places of decimals ; and at the side of each 
page is placed a u Table of proportional parts" by which may 
be found the mantissa of the logarithm of a number containing 
six or seven places of figures : and conversely, if a logarithm 
be given whose mantissa is not contained exactly in the tables, 
the number corresponding to it may be determined to six or 
seven places of figures. 

23. On the construction and use of the tables of 
proportional parts. 

Let m Y and m 2 be the mantissa? of two consecutive integral 
numbers, n and n + 1, which contain five digits each, m the 

mantissa of n + — a number of six digits, the last of which (a) 

is after the decimal point. 



139 

Now since n and n -\ have the same number of integral 

places, their logarithms have the same characteristic ; 

.-. rn-m l = l l0 In + —J -l^n 



a 
n + — 

= i\Q Art. 5. = liQ 

n 



- , nearly, 

Z £ 10 \0n J 



( 1+ 7^) 



by expanding Z 10 ( 1 -\ ) by (ii.) and neglecting the suc- 
ceeding terms as being small compared with the first term. 



Similarly, m 2 - m^ = l 10 (n + 1) - l 1Q n = . — ; 



. . rn — m 1 — \m 2 — m{) — 



Whence m may be found if m 1 , rn 2 , a, be given, — or a may be 
determined when m 19 m 2 , m, are known. 



24. By the Tables, page 147, 

Mantissa of 36633 = m 2 = '5638725 
36632 = m l = '5638606 



.'. m 2 -m l = -0000119 



140 



And in the expression m — m x = (m 2 — m^) — , writing for a the 
numbers 1, 2, 3, 4, 5, 6, 7 5 8, 9 successively, we have 



a 


m 2 —m Y 




1 


1 
•0000119 X — 
10 






or -00000119 


•0000012 nearly 


2 


•00000238 


•0000024 


3 


•00000357 


•0000036 


4 


•00000476 


•0000048 


5 


•00000595 


•0000060 


6 


•00000714 


•0000071 


7 


•00000833 


•0000083 


8 


•00000952 


•0000095 


9 


•00001071 


•0000107 



Now for numbers near 36600 the " dif- 
ference" put down in the Tables is 119, and 
the Table of proportional parts is as in the 
margin. 

We see then that the significant digits 
only of the whole difference and of the 
differences corresponding to the several digits 
are inserted in the table of proportional 
parts. Hence, the following method of 
constructing tables of proportional parts is 
evident : 



119. 


1 


12 


2 


24 


3 


36 


4 


48 


5 


60 


6 


71 


7 


83 


8 


95 


9 


107 



141 



Of the significant part of the whole difference point off 
the last digit as a decimal. Multiply this number by 1, 2, 
3,.., 9 successively, and the whole numbers thus obtained (the 
last digit in the integral part being increased by unity where 
the decimal part is not less than -5) are the significant parts 
of the differences for the digits respectively. 



Thus let the significant part of the whole difference be 156. 



15 



6x1= 15 • 6 = 16 nearly. 

. X 2 = 31-2= 31 

. x 3= 46 • 8 = 47 

. X 4 = 62 • 4 = 62 

. x5= 78-0= 78 



15-6x6= 93-6= 94 nearly. 
... x 7 = 109-2 = 109 
... x 8 = 124-8 = 125 
... X 9 = 140-4= 140. 



By constructing a table of this kind we see that if the 
digit be given the difference is immediately known, or vice 
versa. To avoid the necessity of subtracting in any par- 
ticular case to find the whole difference, there is a line in the 
tables marked at the top with Diff., in which the difference is 
placed opposite to that logarithm at which such difference 
begins. To know the difference therefore in any case, it is 
merely requisite to take the number in this line next above the 
logarithm in question. 



Ex. 1. To find the number whose logarithm is 3*5677766. 



By the tables, p. 147, the mantissa next below the given 
mantissa is that of 1^36963, and the whole difference put 
down is 117' 



142 
Mantissa of the given logarithm = m = -5677766 
Mantissa of l 10 36963 =m l = "5677672 



.*. m — m l = 



94 



By the table of proportional parts to Diff. 117? the dif- 
ference 94 corresponds to the digit 8, — therefore the significant 
part of the number sought is 369638 ; also since the given 
logarithm has 3 for its characteristic, the number required 
is 3696-38. 

Ex. 2. Required the logarithm of 367*654. 

By p. 147., l l0 367'65 = <Z'5654>3*6 
And Diff. being 118, Part for 4 = 47 



.-. Z ]0 367-654 = 2-5654,393. 



25. To find the mantissa of the logarithm of a number 
consisting of seven places of digits. 



Let m Y and m 2 be the mantissae of n and w+1, two successive 

integers of five places each : let M be that of n + 1 

& r 10 100 

which has the same number of integral places as n and n + 1 , 
and also one digit (a) in the place of the tenths, and another 
(b) in that of the hundredths. 

Now since n and n -\ 1 have the same number of 

10 100 

integral places their logarithms have the same characteristic ; 



143 



/ a 6\ , 

M-m^ l l0 1 n + — + —J - / 10 rc 



a b 

n + — + 

10 100 

^io 



M ,+ ;(s + i55)} 



1 1 
/ £ 10 n 

neglecting the terms of the series after the first. 



( a M , 

. — + by (ii.), 

\10 100/ J v } 



Similarly, m 2 — m l = Z 10 (n + 1) - / 10 ?z 

i i 

~ J 6 10 # n 

[a b \ 

.-. M - m 1 = (m 2 - m,) — + 

' VlO 100/ 

= (m 2 - mA — + — { {m 2 - m{) — > . 
V J 10 10 [ V J 10J 

Now (m 2 - m x ) — is the quantity to be added to m l for the 

first additional digit a, Art. 23. ; and (m 2 - m^) — is what 

would have been added had b been the first additional digit : — 
wherefore the quantity to be added for b when it is the 
second additional digit, namely, 



1 f> x H 

10 \ v ' lOj 



is the tenth part of what would have been added had b been the 
first instead of being the second additional figure. 



144 

The following Examples will explain what has been 
said. 

Ex. 1. To find l 10 3684-286. 

By the tables, p. 147, l lQ 3684*2 = 3-56634>32, and Diff. being 
118 the part for a first additional digit 8 is 94, and for a 
first additional digit 6 the part is 71, and therefore the part 

.71 
for a second additional digit 6 is — or 7*1. The operation is 

thus performed; 

Z 10 3684-2 = 3-56634,32 
Part for 8 = 94 

Part for 6= 7*1 



.-. L a 3684-286 = 3'5663533 



Ex. 2. To find the number whose logarithm is 2-5656560. 
2-5656560 
Now 2-5656471 = / 10 367*83, p. 147. 



89 

83 = Part for digit 7- Diff. being 118, 

6 = Part for a second digit 5. 

2-5656560 = Z 10 367*8375. 
and 367*8375 is the number sought. 



145 



26. On the adaptation of formula to logarithmic com- 
putation. 

Multiplication and division being rendered more easy by 
the use of logarithms, while addition and subtraction cannot be 
performed by them, in order to avail ourselves of the ad- 
vantages of logarithms, formulae ought to be so arranged that 
it may not be required to perform the operations of addition 
and subtraction. When this is the case, the formulae are said 
to be adapted to logarithmic computation. 

Thus if a, b, c be given, logarithms cannot be applied to 
determine the value of cos A from the equation, 

b 2 + c 2 - a 2 



cos A 



2bc 



but if from this equation we deduce the formula, 



cos 



v 



S. (S-a) 
be 



where S = -^ 



(a + b + c). 



we can immediately apply logarithms to determine the value 

* A 

OI cos — . 

2 

For a, b, c being given, S and S — a are easily determined, 
and these being known, cos — is determined from the equation 



2 



A 



U cos - = 1 { l l0 [S . (S - a)] - l 10 be} 

= i {hoS+ l l0 (S-a) -l 10 b-l 10 c} 



The two accompanying pages of logarithms are taken from 
Babbage 1 s Tables, the most correct and the best arranged of 
any which have been published. The columns are omitted 
which in those Tables are given to determine the number of 
seconds in an angle containing a given number of degrees, 
minutes, and seconds, and conversely. 

T 



146 



110000000 
2 ! 3010300 
3J4771213 
4,6020600 
56989700 

67781513 

78450980 

8 ! 9030900 

9J9542425 

10 0000000 



0413927 
0791812 
1139434 
1461280 
1760913 

2041200 
2304489 
2552725 
192787536 



20 

21 

22 
23 
24 
25 



27 
28 

29 
30 



31 

32 
33 
34 
35 

36 
31 
38 
39 
40 



3010300 

3222193 
3424227 
3617278 
3802112 
3979400 



264149733 
4313638 
4471580 
4623980 
4771213 



4913617 
5051500 
5185139 
5314789 
5440680 

5563025 
5682017 
5797836 
5910646 
6020600 

41 6127839 

42 6232493 
436334685 
44 6434527 
456*532125 

466627578 

476720979 
486812412 
49,6901961 
506989700 



51 


7075702 


101 


52 


7160033 


102 


53 


7242759 


103 


54 


7323938 


104 


55 


7403627 


105 


56 


7481880 


106 


51 


7558749 


107 


58 


7634280 


108 


59 


7708520 


109 


60 


7781513 


110 


61 


7853298 


111 


62 


7923917 


112 


63 


7993405 


113 


64 


8061800 


114 


65 


8129134 


115 


66 


8195439 


116 


61 


8260748 


117 


68 


8325089 


118 


69 


8388491 


119 


70 


8450980 


120 


71 


8512583 


121 


72 


8573325 


122 


73 


8633229 


123 


74 


8692317 


124 


75 


8750613 


125 


76 


8808136 


126 


77 


8864907 


127 


78 


8920946 


128 


79 


8976271 


129 


80 


9030900 


130 


81 


9084850 


131 


82 


9138139 


132 


83 


9190781 


133 


84 


9242793 


134 


85 


9294189 


135 


86 


9344985 


136 


87 


9395193 


137 


88 


9444827 


138 


89 


9493900 


139 


90 


9542425 


140 


91 


9590414 


141 


92 


9637878 


142 


93 


9684829 


143 


94 


9731279 


144 


95 


9777236 


145 


96 


9822712 


146 


91 


9867717 


147 


98 


9912261 


148 


99 


9956352 


149 


100 


0000000 


150 



0043214 
0086002 
0128372 
0170333, 
02 11 893 

0253059 
0293838 
0334238, 
0374265 
0413927 

0453230 
04921 80 
0530784 
0569049 
0606978 

0644580 
0681859 
0718820 
0755470 
0791812 

0827854 
0863598 
0899051 
0934217 
O969IOO: 

1003705 
1038037 
1072100 
1105897 
1139434 

1172713 
1205739 
1238516 
1271048 
1303338 

1335389 
1367206 
1398791 
1430148 
1461280 

1492191 

1522883 
1553360 
1583625 
1613680 

1643529 
1673173 
1702617 
1731863 
1760913 



151 
152 
153 
154 
155 

156 
157 

158 
159 
160 

161 
162 
163 
164 
165 

166 
167 
168 
169 
170 

171 
172 
173 
174 
175 

176 

177 
178 
179 
180 

181 

182 
183 
184 
185 

186 
187 
188 
189 
190 

191 
192 
193 
194 
195 

196 
197 
198 
199 
200 



17897691 
1818436 
1846914 

1875207 
1903317 

1931246 
1958997 
1986571 
2013971 
2041200 

2068259 
2095150 
2121876 
2148438 
2174839 

2201081 

2227165 
2253093 
2278867 
2304489 

2329961 
2355284 
2380461 
2405492 
2430380 

2455127 
2479733 
2504200 
2528530 
2552725 

2576786 
2600714 
2624511 
2648178 
2671717 

2695129 

2718416 
2741578 
2764618 
2787536 

2810334 
2833012 
2855513 
2878017 
2900346 

2922561 
2944662 
2966652 
2988531 
3010300 



201 
202 
203 
204 
205 

206 
207 
208 
209 
210 

211 
212 
213 
214 
21o 

216 
217 
218 
219 
220 

221 

222 
223 
224 
225 

226 

227 
228 
229 
230 

231 
232 
233 
234 
235 

236 
237 
238 
239 
240 

241 
242 
243 
244 
245 

246 
247 
248 
249 
250 



3031961 
3053514 
3074960 
3096302 
3117539 

3138672 
3159703 
3180633 
3201463 
3222193 

3242825 
3263359 
3283796 
3304138 
3324385 

3344538 
3364>591 
3384>565 
3404441 

3424227 

3443923 
34>63530 
3483049 
3502480 
3521825 

3541084 
3560259 
3579348 
3598355 
3617278 

3636120 
3654880 
3613559 
3692159 
3710679 

3729120 
3747483 
3765770 
3783979 
3802112 

3820170 
3838154 
3856063 
3873898 
3891661 

3909351 
3926970 
3944517 
3961993 
3979400 



147 



Num. 



3650 



9 
3660 



5622929; 304S 
4237 
5427 
6616 
7804 

8993 
0181 
1368 
2556 
3743 



4118 
5308 
6497 
7685 

8874 
o630062 
1250 
2437 
3624 



3167 
4356 
5546 



3670 



4811 
5997 
7183 
8369 



4930 
6116 
7302 

8488 



5048 
6235 
7421 
8606 
955519673, 9792 



3286| 3405 
4475! 4594 
5664! 5783 
6734' 6853 6972 
7923^8042 8161 

9111 9230 9349 
0299 0418 0537 
14871 1606 1725 
2674 2793 2912 
3861 3980 4099 



5640740, 0858 
1925 2043 
3109 3228 
4293 4412 
5477 5595 

6661 6779 
7844! 7962 



Log. 562. N. 365. 
6 7 8 9 Diff. 



5167 

6353 
7539 
8725 
9910 



0977 IO95: 
2162 2280! 



7 
8 
9 

3680 



9 

3690 
1 

2 
3 
4 

5 
6 

7 
8 



9027 
5650209 
1392 

2573 
3755 
4936 
6117 

7298 

8478 
9658 
5660838 
2017! 
3196] 

4375 
5553 
6731 
7909 
9087 

5670264 
1440 
26171 
3793! 
4969; 

6l44| 
7320; 
8495, 
9669 { 
5680843 



9145 
0328 
1510 

2692 
3873 
5054 
6235 
7416 

8596 
9776 
0956 
2135 
3314 

4493 
5671 
6849 
8027 
9204 

0381 
1558 
2735 
3911 
5086 

6262 
7437 
8612 
9787 
09fa 

1 



3346 
4530 
5714 

689 

8080 

9263 

0446 

1628 

2810 
3991 
5173 



8714 
9894 
1074 
2253 
3432 

4611 

5789 
6967 
8145 
9322 

0499 
1676 
2852 
4028 
5204 

6379 
7555 
8729 
9904 
1078 



5285 
6472 
7658 
8843 
o029 

1214 

2398 

3583 

476 

5951 

7134 
8317 
9500 
0682 
1 

3046 
4228 
5409 
6590 
7770 

8950 
o!30 
1310 
2489 
3668 

4846 
6025 
7203 
8380 
9557 

0617 0734 
1793 19H 



3464 
4648 
5832 

7016 

8199 
9382 
0564 
1746 

2928 
4109 
5291 



6353 6471 
7534 7652 



8832 
o012 
1192 
2371 
3550 
I 
4728 

5907 
7085 
8262 
9440 



3524 3642 
4713)4832 
5902, 6021 
70911 7210 
8280 8398 

9468 9587 
0656; 0775 
1843 
3031 
4218 



5404 
6590 
7776 
8962 
ol47 

1332 
2517 
3701 

4885 
6069 

7252 
8435 
9618 



1962 
3149 
4336 

5523 
6709 
7895 
9081 
o266 

1451 

2635 
3820 
5004 
6187 

7371 
8554 
9736 



3761 
4951 
6140 



3880 3999H9 



5070 
6259 



7329J 7448 
8517 8636 

970519824 



5189 
6378 
7567 
8755 



9943 
0893 1012 1131 
2081 2200 



3268|3387 
4455 4574 



0800 0919 
1983 2101 

3164 3282 
4346 4464 



5641 
6828 
8013 
9199 
c384 

1569 
2754 
3938 
5122 
6306 



5760 
6946 
8132 
9318 
o503 

1688 
2872 
4056 
5240] 
6424 

7607 
8790 
9973 
1155 



119 

11 12 

2 24 

3 36 

4 48 
5| 60 
G 71 
7l 83 

8 95 

9 107 



2970 

4146 
5322 

6497 
7672 
8847 
c021 
1196 



3087 
4263 
5439 

6615 
7790 
8964 
ol 39 
1313 



3 4 



5527 
6708 

7888 

9068 
o248 
1428 
2607 
3786 

4964 
6142 
7320 
8498 
9675 



5645 
6826 
8006 



7489! 
86721 
9855 
1037 
2219 

3401 3519 

4582 4700 
5763 5881 
6944j 7062 
8124 8242 



9186(930419422 
o366 o484 o602 
1545 1663 1781 



2318 
3505 
4692 

5879 
7065 
8251 
9436 
o621 

1806 
2991 
4175 
5359 
654,2 

7726 



118 



2725.2843 
3903 4021 

5082 5200 
6260, 6378 
7438! 7556 
8616J 8733 
9793 9911 



0852 0970] 1087 

2029J 21461 2264 

3323; 3440 



3205 
4381 
5557 

6732 
7907 
9082 
o256 
1430 



4499 4616 
5674 5792 



6850 
8025 
9199 
o374 
154S 



6967 
8142 
9317 
o491 
1665 



2960 
4139 

5318 
6496 
7674 
8851 
o028 

1205 

2382 
3558 
4734 
5909 

7085 
8260 
9434 
06O8 
1782 



c091 
1273 
2455 

3637 
4818 
5999 
7180 
8360 

9540 
o720 

1899 
3078 
4257 

5435 
6614 
7791 
8969 
ol46 

1323 
2499 
3675 
4851 
6027 



7202 
8377 
9552 
o726 
1900 



6 7 



118 
1 12 



94 

9106 



117 



117 

11 12 

2 23 

3 35 

4 47 

5 59 
fi 70 
7; fi2 
8 94 
9105 



APPENDIX II. 



ON THE CONSTRUCTION AND USE OF TABLES OF GONIOMETRIC 
FUNCTIONS. 



1. In applying trigonometrical formulae it is convenient 
to have the values of the trigonometrical functions of angles 
registered in tables, and so to avoid the necessity of computing 
them when wanted in any particular case. 

But because in very small angles the sines and tangents 
are exceedingly small, and contain after the decimal point two 
or three places of cyphers before we come to the significant 
figures, the goniometrical functions registered in the tables 
have had the decimal point moved four places to the right, 
that is, have been multiplied by 10 4 or 10000. The Tables so 
formed are called, " Tables of natural Trigonometric functions 
of angles," or " Tables of goniometric functions*." 

* The tables of goniometric functions are sometimes said to be " calculated to a 
radius of 10000 ; " the meaning of this expression we will explain. 

If with C as center and radius CA an arc AB be drawn, and 
BN, AT he perpendicular to CA, the lines BN, CN, A T, 
CT, AN respectively, are sometimes defined to be the sine, 
cosine, tangent, secant, and versed sine, of the angle ACB to 
the radius CA, — or the sine, cosine, tangent, secant, and 
versed sine, of the arc AB. 

c 

The expression " to the radius CA " is absolutely necessary to the definitions, 
because the lines BN, AT.... evidently depend on the magnitude of A C as well as 
upon the magnitude of the angle ACB',— in fact, the angle ACB continuing of the 
same magnitude, they vary directly as CA. 

Now by the definition that we have used of the sine, sin ACB = 7777 =77^ T l 

C x> C A 

.-. 10000 x sin ACB = BN. 10 J}^ . 
CA 

But 10000 . sin ACB is the tabular sine of ACB; 

.: tab. sin ACB = B N. ^^=B N, if CA = 10000. 
C A 

Wherefore the tab. sine of ACB expresses the magnitude of the line BN, (the 
sine of ACB to the radius CA), the magnitude of CA being represented by 10000. 

Similarly the tab. cosine, tab. tangent,... of ACB express the magnitudes of the 
lines CN, AT, ...the magnitude of CA being 10000. 




149 

2. To find the values of the sines and cosines of every 
angle less than 90°. 

By Art. 35. (2), 2 (cos Af = 1 + cos 2 A, 
or cos A - V^(l + cos 2 A); 

.-. cos— = y/ 1 (l + cos A), 

2 



cos — = \/ i f 1 + cos — ) ; and so on. 

2 2 z \ 2/ 

Now if A = 30°, cos2j = -|. Art. 31. (2). And having 
performed the successive operations indicated above eleven 
times, we have 

30° 
cos — , or cos 52"-734375, = '99999996782 ; 



30' 



.-. sin 



|0 /if so°\ 2 ) 

- = ^/ J ] _ I cos — > = -000255663462. 



30° 30 . 6o' 
3. Now since -— = — '- — , and the sines of small angles 

are as the number of seconds, and therefore as the number 
of minutes, they contain, Art. 103. Cor. ; 

. , . 30° , 30.60' 
.-. sml : sin — .:: V : — ; 

. , -2 U . 30° 

.-. sin 1 = — — sin — - 

30.60 2 n 

2 8 . .30° 

= sin — 

15,15 2 11 

= x -000255663462 

225 

= -000290888204. 

And from sin l' we may thus determine sin 30" ; 

sin 30"= sin |'= \ {^/(i + sin l') - ^/(l - sin l')}, Art. 37- 



150 

4. Sin l' and sin 30" being known, the sines of l', 2', 3' ... 
to 60' (or 1°) may be determined. 

For sin ( A + B) + sin (J - 5) = 2 sin J . cos 5 

= 2 sin J Jl - 2 (sin — ] 1 

2 sin — ) , 

/ 7?\ 2 

\ sin {A + 5) = sin A 4- { sin J - sin (J - 5) } - sin A f 2 sin — ) ? 

If in this formula 2? = l', and we write l', 2', 3' 

successively for A, we have 



sin 2' = sin l' + (sin l' - sin 0') - sin l' (2 sin 3u") 2 , 

sin 3' = sin 2' + (sin 2' - sin l') - sin 2' (2 sin 30") 2 , 

sin 4' = sin 3' + (sin 3' - sin 2') - sin 3' (2 sin 3u") 2 , &c. 

This method is not very laborious : — in the last example, 
sin 3' and sin 3 r — sin 2' are known from the two preceding 
lines ; and the only labour is in multiplying sin 3' by the factor 
(2 sin 30") 2 which is common to all the terms. 

Cor. If in the above formula 5=1° and we write 

1°, 2°, 3° successively for A, we can determine the sines 

of 2 n , 3°, 4° 

5. The sines of angles zip to 60° having been calculated, 
those of angles between 60° and 90° may be thus determined. 

Sin (60° + A) - sin (6o°- A) = 2 cos 60°. sin A. 

And cos 60° = J; 

.-. sin (60° + A) = sin A + sin (60° - A). 

And if ^4 increase by l' from 0° to 30°, this formula will by 
addition merely give us the sines of angles from 60° to 90°. 



151 



6. The sines of angles up to 90° having been determined, 
their cosines are also known. 

For cos A = sin (90° - A). 
Thus cos 25° = sin (90° - 25°) = sin 65°, 
cos 72° = sin (90° - 72°) = sin 18°. 

7- The tangents, cotangents, secants, and cosecants can 
be determined from the sines and cosines. 

sin A cos A 

r or tan A — , cot A — — , 

cos A sin A 

A X A l 

sec A = , cosec A - 



cos A sin A 



8. Since Art. 16. p. 8, sin A = cos (90 - A), 

Art. 25. p. 16, tan A = cot (90° - A), 

sec ^ = cosec (90 — ^f), 

the sines, tangents, and secants of angles from 45° to 90° are the 
same respectively as the cosines, cotangents, and cosecants of 
angles from 45° to 0°. Wherefore it is not necessary to carry 
the tables further than to the angle 45°, 

Thus cos 72°, 20' = sin (90° - 72°, 20') = sin 17°, 40', 

sin 72°, 20' = cos 17°, 40', 

tan 72°, 20 r = cot 17°, 40', 

sec 72°, 20' = cosec 17°, 40'. 

At the bottom of the page containing the functions of 
angles from 17° to 18° is placed the angle 72°, and the column 
which at the top of the page is marked to indicate the sines 
of angles from 17° to 18°, is marked at the bottom to shew 
the cosines of angles from 72° to 73° ; and so for the other 
functions. See page 161. 



9. 



152 



On Formulae of Verification. 



Since the trigonometrical functions of angles are deter- 
mined successively from one another, one error will affect every 
successive result. As checks against the possibility of errors, 
several formulae (of verification as they are called) are used to 
examine the accuracy of the results, and the values registered 
in the tables are presumed to be correct if they satisfy these 
formulae. 

The following are the principal formulae of verification. 

(1) Sin A = 1 {«s/l+$m2A +\/l -sin 2 J}. Art. 37. 

(2) Cos A = \ {vising J -\A -sin2j}. 



Again, cos 36° = ^ 5 + l , cos 72° = — — -; Art. 49- (3) (l). 

° 4 4 

/— 
.-. sin(36°+^)-sin(36°-l) = 2cos36°.sinJ= — . sin A, 

„ . d y/$ - 1 . t 
and sin (72° + A) - sin (72° - A) =2 cos 72° . sin A = . sin A ; 

Art. 43. (2). 
.-. by subtraction we have, 

sin (36° + J) + sin (72°- A) -sin (36°- J) - sin (72°+ A) = sin J ; 

(3) .'. sin .4 + sin (36°- J) + sin (72°+^) = sin (36°+ J) + sin (72°- 4), 

which is Enter's formula. 



Again, sin 54° = ^l£jti , sin 18° = ^ & ; Art. 49. (3) (2). 



and 2 sin 54° cos A = sin (54° + A) + sin (54° - A) 
2 sin 18° cos A = sin (18° + A) + sin 



in ^54" - A) | 

in (18°- J) J 

Art. 43. (1). 



153 

(4) .-. cos A, or sin (90° - J), 

= sin (54° + A) + sin (54° - J) - sin (18° + A) - sin (18° - A), 

which is Legendre's formula. 

This formula might have been deduced from Euler's by 
writing 90° - A for A. 



Ex. To exemplify the use of these formulae, we have 
from (4) by making A — 13°, 

cos 13° = sin 67° + sin. 41° - sin 31° - sin 5°. 

Now the tables give for the quantities in the second member of 
the equation 

9205-049 + 6560-590 - 5150-381 - 871 '557, 

which = 9743-701, 

the quantity given by the tables as the cosine of 13°. 

Since therefore these quantities satisfy the relation which 
we know ought to exist between the sines of 67°, 41°, 31°, 5°, 
and the cosine of 13°, we conclude that the values of these 
natural goniometric functions are rightly given by the tables. 

10. The values of the goniometric functions having been 
thus calculated, multiplied by 10000, verified, and registered in 
tables, are called Ci Tables of natural sines, cosines, tangents, 
cotangents, secants and cosecants." To find the real gonio- 
metric functions from these tables, we have merely to divide 
the tabular numbers by 10000, that is, to remove the decimal 
point four places to the left. 

11. Since the natural trigonometric functions of angles 
registered in the tables are calculated only for angles which 
contain degrees and minutes, it is necessary to establish a rule 
by which we may be able to determine the natural gonio- 
metric functions of angles which contain seconds in addition 
to degrees and minutes. 

U 



154 

The increments of each of the tabular goniometric func- 
tions, except in some particular cases, vary as the increments 
of the angle. 

Let A be an angle containing a certain number of degrees 
and minutes. 

Let Si = sin A, 

S 2 = sin (A + l') = sin (A + 60"), 
S = sin (A + §A"). 

Now by Art. 51. Cor., except when A = (2n + l) 90° nearly, 
S 2 — Si = cos A . sin 6o", 
S - Si = cos A . sin $A" ; 
S - Si sin $A" 



" S 2 -Si sin 60" 



= Art. 102. Cor. 

60 

10 4 .ff -10 4 . Si $A 
""' io 4 . tf 2 - io 4 . Si ~~~6o" 

And lO 4 .^, 10 4 .*S r 2 , IO 4 .*?, are the tabular sines of the angles 
A, A + 60", J + SyT. 

Wherefore, if s 1? s 2 , s represent the tabular sines of these 
angles, we have 

5 - Sj $ A 
s 2 — Si 60 ' 

*—*-*'•*£ (I.) 

BA= s ~ Sl (2.) 



St! - Si\ 

V 60 ) 



Whence, since s 2 — s Y is known from the Tables, 

by (l) s — Si may be found, if $ A be given ; 
by (2) SA may be found, if s - s, be given. 



155 

Similarly, if c 15 c 2 ,c; <ri,o- 2 , <r; t 19 t 2 ,t; be the tabular 
cosines, secants, and tangents of the angles 

A, A + 60", A + U", 

respectively ; we have 

Unless A = 2n.90° nearly, Art. 52. Cor., 

c — c x sin A . sin $A" sin cA" §A 
c 2 - Cj sin A . sin 6o" sin 6o" 60 ' 

Unless A =n.90° nearly. Art. 53. Cor. 2., 

cr -(Ti tan A . sec A . tan $ A" tan$A" cA _ 

L = — = = ; Art. 108. Cor. 

<r 2 - o-j tan A . sec A . tan 60 tan 6o" 60 

Unless J = (2n + l) • 90° nearly, Art. 54. Cor., 

t-t, _ (sec J) 2 .tang,A" _ tan cA" _ %A 
t 2 -t Y ~ (sec Af . tan 60" ~ tan 60" ~ 60 ' 

Hence we collect, that the increment of the goniometric 
function varies as the increment of the angle, 

For the sine, except when A = (%n+ 1) 90° nearly. 

cosine, A= 2n.90° 

secant, A= n.90° 

tangent, A = (2n + l) 90° ....... 

12, After the columns of natural sines, cosines in 

the Tables, see p. l6l, there is a line headed " ND for l" 9 V 
that is the Natural Difference for l"; — the meaning and use 
of this line we will now explain. 



156 



In Art. 11. (I) if 3 A" become l", is the difference 

v ; 60 

for one second ; and if this quantity be calculated and regis- 
tered, the labour of computing s — s x when 8 A is given, is 
reduced to multiplying this registered quantity by 8 A; or 
if hA be required from s — s 19 we have merely to divide s — s x 

s — s 
by the registered quantity — - — -. Art. 11. (l) and (2). 



Thus, required the tab. sine of 17°, 0', 12". 
t . sin 17°, l' = s 2 = 2926-499 p. l6l. 
t. sin 17°, 0' = *, = 2923717 



s 2 - s, = 2-782 
s 2 - s x 2-782 



•046.36 ; 
60 60 

.\ s - s } = -04636 x ]2 = -55632; 

.'. s a «j + -55632 = 2924-27332 

= 2924-273 nearly. 

13. It is to be observed, p. 161, that the registered u ND 
for l"" for the natural sines of angles between 17° and 17°, 4' is 
not -04636, (as might have 'been expected from what was said 
in the last Article), but 46*36, — the decimal point being 
removed three places to the right. Now if 46-36 be multiplied 
by 12, the result is 556'32, or 556 nearly ; and if we place 
this quantity so that the units' digit comes under the last 
figure of t . sine 17°, 0' and the figures be added vertically, 
the same value of t . sine 17°, 0', 12" is obtained as before. 



157 

The reason therefore of removing the decimal point of 

~ three places to the right is obvious; — it is to avoid 

60 r ° 

the necessity of printing cyphers. 

What has here been said of the tabular sines is equally 
applicable to any other tabular goniometric function. 

14. Ex. 1. Required the tabular sine of 17°, 3', 8", 
" ND for l"" by the Tables, p. l6l, is 46" 36 ; 

4,6-36 
8 



370-88 = 371 nearly. 

And t . sin 17°, 3' = 2932-06 1 



.-. t.sm 17°, S , 8 = 2932-432. 

Ex. 2. Required the angle whose t . sine is 2996*684, 

The next less t . sine is that of 17°, 26', and the ND for 1 ; 
is 46'25. 

2996*684 
£.sin 17% 26' = 2995'959 



* -725; 

725" 
46-25 ' 



("725 \ 
i. e. — — - , = 15"-675 = 15" -68 nearly. 
•04625/ J 



And the angle required is 17°, 26', 15"'6S nearly. 



158 

Ex. 3. Required the tabular cosine of 28°, 35', 36". 
t. cos 28°, 35' = 878 1*222, and "N.D for l"" is 23'2 ; 
.-. N.D for 36" = 23-2 x 36 = 835'2. 

But since as the angle increases the t . cosine decreases, we 
must subtract this quantity from £.cos28°, 35', in order to 
obtain t . cos 28°, 35', 36" ; 

.'. t cos 28°, 35', 36" = 8780-387- 

It may here be remarked, that the difference for additional 
seconds must be added for tabular sines, tangents, and secants, 
Arts. 51, 54, 53, — and subtracted for tabular cosines, Art. 52, 
cotangents, and cosecants. 

15. In some cases, although the increment of the gonio- 
metrical function vary as the increment of the angle, yet 
this principle cannot be used to determine the increment of 
the angle, owing to the tables being calculated to an insuf- 
ficient number of decimal places. 

Let 8 19 s 2 , s, be any, the same, tabular functions of the 
angles A, A + 6o", A + $A ff respectively; and let it be true 
that 

,s s 9 — S-\ 
s - s, = 6 A 



60 

The common Tables are calculated to seven places of 
figures ; if therefore A be of such magnitude that s 2 — Si be 
not contained, or it be but just contained in seven places, 
then s — $! , which is less than s 2 — s Y , will not be contained in 
seven places ; and therefore s - s x cannot be accurately de- 
termined. 

Thus t . sin 89°, 45' = 9999*905, and " ND for l" " is -20. 
Let n be the number of seconds which increases the last 
figure of this t . sine by unity, then 

»" i „ i" ,, 



159 

The tabular sines therefore of all angles between 89°, 45' and 
89°, 45', 5" will be the same for seven places of figures, and 
we cannot determine which of the angles from 89°, 45' to 
89°, 45', 4" is to be taken for that whose tabular sine is 9999'905. 

It is also evident that if the angle 89° 45' were diminished 
by 4", the tabular sine would have the same seven places 
of figures, and this consideration renders the angle whose 
t . sine is 9999-905 yet more indeterminate. 



unity, n is greater than unity ; that is, we cannot from its 
given function determine an angle accurately to seconds unless 
the " N . D for l" " amount to unity at least. By referring 
to the Tables, it will be seen that for sines, the "ND for l"" 
becomes less than unity when the angle exceeds 88°, 50'. 

16. From what has been said in Arts. 11, 14, it is obvious 
that to determine the additional seconds in any particular case, 
it must be true that, 

I. The increment of the function does in that case vary 
as the increment of the angle. Art. 11. 

II. The increment of the function corresponding to l", 
or " ND for l"," i s > not in that case very small. Art. 14. 

Also, when 1. is true, it is evident that the greater " ND 
for l"" is, the more accurately can the angle be determined 
from its given natural function. 

17- The magnitude of an angle which is less than 45° 
can be more accurately determined from its sine than from 
its cosine. 

For A sin A = cos A . sin § A = cos A . sin l", if $A = l". 
A cos A - - sin A . sin $A — - sin A . sinl'', 



160 
But if A be < 45°, sin A is < cos J; 

.'. A sin J for l" > A cos A for l"; 
> "AM) for l"" for sin A > "N.D for l"" for cos A. 

Hence (Art. 15.) A can in this case be more accurately 
determined when its sine is given than when its cosine is 
given. 

Similarly, it may be shewn that if A be > 45° and < 90°, it 
may be more accurately determined from its cosine than from 
its sine. 



A page from Sherwin's Logarithmic Tables, corrected in 
some places, is here subjoined, the natural cosines, tangents, 
secants, and cosecants being omitted. 



161 



17 Degrees. 



M\N. Sine, 



2923717 

12926-499 

22929-280 

32932061 

42934-842 

5*2937-623 

62940-403 

7(2943-183 

8,2945-963 ' 

9,2948743 

10 2951522 
li;2954-302 53 
12 2957-081 o 
13'2959-859 "* 
142962-638 

152965-416; 
162968-194! « 

17:2970-971! g 

18 2973-749 ^ 
19J2976-526 

20.2979-303 
2l'2982-079 co 
22'2984-856 £ 
232987-632 ^ 
242990-408 
25 2993-184 
262995-959 ^ 
2712998-734! <* 
28:3001-509 **■ 
293004-284 

303007-058 

31 3009-832 

32 3012-606 
~~ 3015-380 

3018153 



3020-926 

3023-699 

3026-471 

38 3029-244 



3032-016 
3034-788 
3037559 
3040-331 



43 3043102 

44 



M 



3045-872 

3048-643 
3051-413 
3054-183 
3056-953 
3059723 

50 3062-492 
3065-261 

52 3068030 
3070798 
3073-566 



3076-334 
3079102 



54 

5813084-636 

59 3087-403 

60 3090-170 



L. Sine. 



9-4659353 
9-4663483 
9-4667609 
9-4671730 
9-4675848 

9-4679960 
9-4684069 
9-4688173 
9-4692273 
9-4696369 
9-4700461 
9-4704548 
9-4708631 
9-4712710 
9-4716785 

9-4720856 

9-4724922 

•4728985 

•4733043 

9-4737097 

4741146 

9-4745192 

9-4749234 

9-4753271 

9-4757304 

9-4761334 
9-4765359 
9-4769380 
9-4773396 
9-4777409 
9-4781418 
9-4785423 
9-4789423 
9-4793420 

4797412 
9-4801401 

4805385 
9-4809366 
9-4813342 
9-4817315 

4821283 
9-4825248 
9-4829208 
9-4833165 
9-4837117 
9-4841066 
9-4845010 
9-4848951 
9-4852888 
9-4856820 

4860749 
9-4864674 
9-4868595 
94872512 
9-4876426 

9-4880335 
4884240 
9-4888142 
94892040 
9-4895934 
9-4899824 



Co-sines. 



Biff. 
1 sec. 



68-833 
68766 
68-683 
68-633 
68-533 

68-483 
68-400 
68-333 
68-266 
68-200 

68-116 
68-050 
67-983 
67-916 
67-850 

67-766 
67-716 
67-633 
67-566 
67*483 

67-433 

67-366 

67-283 

67-216 

67-166 

67-083 

67-016 

66-933 

66-883 

66-816 

66-750 

66-666 

66-61 

66-533 

66-483 

66-400 
66-350 
66-266 
66-216 
66-133 

66-083 
66-000 
65-950 
65-866 
65-816 
65-733 
65-683 
65-616 
65-533 
65-483 
65-416 
65-350 
65-283 
65-233 
65-150 

65-083 
65-003 
64-966 
64-900 
64-833 



Co-secants.l L. Tan 



0-5340647,9-4853390 
0-53365l7i9-4857907 
0-5332391s9-4862419 
0-53282709-4866928 
0-5324152 9-4871433 

0'5320040'9-4875933 
0-5315931,9-4880430 
0-531182719-4884924 
0-53077279-4889413 
0-53036319-4893898 
0-5299539j9-4898380 
0-5295452:9-4902858 
0-5291369 9-4907332 
0-52872909-4911802 
0-5283215 9-4916269 



•4920731 
•4925190 
•4929646 
•4934097 
•4938545 

4942988 
4947429 
4951865 
4956298 
4960727 
4965152 
4969574 
4973991 
4978406 
4982816 



0-5279144 
0-5275078 
0-5271015 
0-5266957 
0-5262903 

0-5258854 9 

0-5254808 

0-52507669- 

0-52467299' 

0-52426969- 

0-52386669- 
0-52346419' 
0- 523062019- 
0-52266049- 
0-52225919- 

0-5218582i9-4987223 
0-52145779-4991626 
0-52105779-4996026 
0-5206580 9-5000422 
0-520258819-5004814 
5009203 
5013588 
5017969 
•5022347 
•5026721 

•5031092 
•5035459 
■5039822 
•5044182 
•5048538 

0-5158934 9-5052891 
•5057240 
■5061586 
•5065928 
•5070267 

0-5139251 9-5074602 
5078933 
5083261 



0-51985999 
05194615 
0-5190634 9 
0-5186658 
0-5182685 

0-5178717 
0-5174752 
0-5170792 
0-5166835 
0-5162883 



0-5154990 
0-5151049 9 
0-5147112 
0-5143180 9 



0-5135326 
0-5131405 9 
0-5127488 9' 
0-5123574 9 

05119665 
0-5115760 9 
0-5111858 
0-5107960 9 
0-5104066 9 
0-5100176 9 



L. Sec. 



5091907 
5096224 
5100539 
5104849 
5109156 
5113460 
5117760 



Co-tans. 



Biff. 
1 sec. 



L. Sec. 



f ~— 100194037 
\\ 283 10-0194423 
75 200j 10 . 0194810 

75 083 l0 . l95585 

It S 1001 96361 
^: 900 J10-0196750 
74 816 10 . 197140 
74 750100197529 

I? 3100197919 
74fi33 ]0 . 0198310 

74;f66 100198701 
It An 100199092 
74*450 100199484 

JiJ? 10-0199876 

74 316 100200268 
74-260 100200661 
74;|83 100201054 
7;iS 10 '0 20 1448 
^.10-0201842 
74 01610-0202236 
7<tS 100202631 
H qY* 10-0203027 
16 »}6 10-0203422 

^W " 0203818 
<{ 'JO 10-0204215 

?o So 100204612 
H fnn 100205009 
73'S 10 '0 205 407 
L.000 10-0205805 
1% S 10-0206204 
i-i'Hi 100206602 
%f 6 100207002 
^ 2 00 100207401 

7o. nflq 100207802 
jg 083 10-0208202 
liala 10-0208603 
( 2 ; 9 66 100209004 
i 2 9 00 10-0209406 

gS 100209808 
( 2 783 10-0210211 
7 2 716 100210614 
7 2 666 100211017 
7 2 600 10 . 0211421 

;'^ 10-0211825 
7 2 483i . 0212230 

11 g3 10-0212635 
7 2 366 100213040 
72-316 10-0213446 

;;^? 10-0213852 
7 2 Jg 100214259 
72 133 100214666 
7 2 083 100215073 
7? SIS 10-0215481 
7.^10-0215889 
71916J0-0216298 
71833 io-0216707 
71/83 100217117 
71 733 100217526 
7166b 100217937 



Cosecants 



Biff. 
1 sec 



6433 

6-450 

450 

466 
6-466 
6-483 
6-500 
6-500 
6-500 

6-516 
6-516 
6-516 
6-533 
6-533 
6-533 

550 
6-550 
6-566 
6-566 
566 
583 
6- 

•600 

•600 
6-616 
6-616 
6-616 
6633 
6-633 
6- 

•633 
6-666 
6-650 
6-683 

6-666 
6-683 
6-683 
6-700 
700 

716 
6-716 
6-716 

733 
6733 

6-750 
6-750 

•750 
6-766 
6-766 
6-783 
6-783 

783 
6-800 
6-800 
6-816 
6-816 
6-833 
6816 

850 



Co -sines. M 



•980596350 
9-9805577 59 
9-9805190 
9-9804803 
99804415 

9-9804027 

" 9803639 

99803250 

9-9802860 

9-9802471 

9-9802081 

9801690 

9801299 

9800908 

9-9800516 



9800124 45 
9-979973244 

•9799339 43 
9-979894642 

•979855241 

9-979815840 
9-979776439 
0-9797369,38 
1-979697337 
1-979657836 

1-979618235 
1-979578534 
1-979538833 
1-979499132 
1-979459331 

J-9794195I30 
9-979379629 
0-979339828 
1-9792998^7 
••979259926 

1-979219825 

99791798|24 

9791397123 

•9790996 22 

9-9790594^21 

9790192,20 



9789386;18 
9788983)17 
978857916 

978817515 

978777014 

978736513 

978696012 

9-978655411 

•9786148;i0 

•9785741 

•9785334 

9-9784927 

9-9784519 

9 9784111 

9-9783702 

9783293 

9-9782883 

9 9782474 

•9782063 



L. Sine. M 
I 



72 Degrees. 
X 



APPENDIX III. 



ON THE LOGARITHMIC TABLES OF GONIOMETRIC FUNCTIONS. 



1. When the trigonometric functions of angles are de- 
termined, their logarithms may be found from the tables of 
the logarithms of numbers. There are however methods by 
which the logarithms of the goniometric functions can be found 
independently. 

2. To find 1 10 sin 0, sin 6 not being given. 

Si ,e = e(l- 6 ^ (l-£) (l-^) Art. ,28. p.,21. 

Making = — . - , and taking the logarithms of both sides 
of this equation, we have 

, . (m ir\ fm tt\ l m 2 \ / m 2 \ 

tr\ , (2 2 n~ — mr 



2 / V 2'n 



\n 

\ im 2 m 4 1 m 6 \ 

"~£l0 lo^ + ^*o^ + ^'6V + '") 
\ I m 2 m 4 /w 6 \ 

^0 \¥n 1 + * ' 8V + 3 • ^ + • • • J 



-&c 



(7W, 2 \ / 771 2 \ 

1 " iw) ' /l0 1 1 " 6V j ' &c> being ex P anded b ? ("•)> P* 1S2 ' J 



lt$ 



Now J 10 (_.~] « / 10 m + Z 10 7r-/ 10 w -l 1Q 2, 
and Z 10 — a = ^ 10 1 (2 t* + m) (2rc - m)} - l l0 2 2 n* 

A 7b" 

- li (2n + in) + l 10 (2n- m) - 2l l0 2 -2l 10 n; 
: /igsm — . - \=l 10 m+l 1Q (2n+<m)+l lQ (2n-m)-3l l0 n--3l l $+l lQ 7r 

\7t ii j 



1 /i i i \ ffl 6 

+ * Vi^ + 6^ + 8^ + "V *^ 



l £ 10 



t + &c. 



And by giving m and w different values, the logarithmic sines 
of all angles may be found by this formula. 

3. In a like manner from the series 
2 2 2 



a ( 2<W\ ( 2'6<\ ( 2<W\ 
we get the formula 

r /111 "i a*? 

/111 \ W 4 

j /l 1 1 \ m 6 

+ 3 ^6 + ^ + ^ + -J ^ 



122. 



Z £ 10 



+ &c. 



164 



4. The logarithms of the sines and cosines having been 
thus determined, those of 

_ sin 1 cos . 1 

tan = , sec = , cot 6 = -; — - , cosec 9 = — — - , 

cos 6 cos sm 6 sin 6 

may be severally found. 

5. Since all sines and cosines are, generally, less than 
unity, their logarithms to base 10 are negative. In order to 
avoid the inconvenience of printing negative characteristics, the 
logarithms to base 10 of all goniometrical functions have been 
increased by 1 0, and the resulting numbers being registered are 
called " Tables of logarithmic sines, cosines, Sec." 

Hence if the tabular logarithmic function of an angle be 
given, by subtracting 10 from it we get the real logarithm of 
the function. 

These tabular logarithmic functions we shall indicate by 
the letter L ; thus the tabular logarithmic sine of A, or 
10+ l l0 sin A, will be written L sin A. 

6. The increment of a tabular logarithmic function of 
an angle varies, except in certain cases, as the increment of 
the angle. 

Let the function be the sine, and let A receive the incre- 
ments SA" and 60" successively. 

'. , * a jff, . A sin ( J -f §A") - sin A) 
Then sm (A + $A) = sin A \l + - — y— ± j 



sin A \ 1 + 



sin A 

cos A . sin $A" 



! 



sin A 
unless A = ( c 2n + 1). 90° nearly ; Art. 51. Cor. 2. 

.-. l ]0 sin (A + $A") as l l0 sin A + l 10 (l + cot A . sin $ A") ; 

.•, { 10+/ 10 *m(A+§A") \ - { 10+ J w sin ^} *ho +cot^.sin $A") ; 



165 
. . L sin (A + 8 A") - L sin A 
= - 1 -.{coU.sin^ ,, -i(coU) 2 .(sin^ , ') 2 +...} App.i.(ii). p.132, 

= . cot A . sin 8 A", 

l t 10 

neglecting the higher powers of cot A . sin 8A" 9 which may be 
done unless A = 2n . 90° nearly. 

And writing 6o" for 8 A" in this equation, it becomes 

L sin (A + 6o") - L sin A = - — . cot J . sin 6o"; 

L sin (A + S A") - L sin A sin S A" 8 A nn _ 

.-. 2 _£ ^ = — . Art. 102. Cor. 

L sin (J + 60") - L sin J sin 6o" 60 

And in a manner exactly similar it may be shewn that for 
any function of an angle the increment of tab. logarithm varies 
as the increment of the angle. 

It is to be observed that unless in the selection of a formula 
the principles i and n of Appendix ii. 15. be carefully borne 
in mind, the increments of angles cannot be accurately de- 
termined from the increments of the tabular logarithmic 
functions, or vice versa. 

7- To explain the meaning and use of the columns of 
differences for one second (Diff. l"), which are placed after 
the columns of logarithmic functions 

If $ A" become l", the equation of Art. 6. becomes, 

L sin (A + l") - L sin A = \L (sin A + 6o") -L sin A} ~ , 

which is the difference of L sin A for one second. 

Now, if this quantity be computed and registered, when 8 A 
is given L (sin A + 8 A") —LsinA may be determined by 
merely multiplying this registered difference by 8A 9 — and 
when L sin (A + 8 A") is given, 8 A may be found by dividing 
L sin ( A + 8 A") - L sin A by this difference. For 



166 



% 4 
L sin {A + $A") -LsinJ = {L sin (A + 60") - Z, sin A] ~ 

L sin (A + 60") — Z, sin A 



60 



.$A, 



ft Z, sm ( J + $J ) — L sin ^f 

and dJ= — — : — — --r--7: r^. — 7. 

Z, sin {A + 60 ) - Z, sin J 

60 
Thus, Z, sin 17°, l' = 9'^6634>83 
£ sin 17° =9*4659353 

.*. L sin 17° l'-L sin 17°= "0004130 

4130 ,.,'-'-, • 1 -i 

Now = 68-833, which is the quantity put down in the 

tables as the difference for one second to sin 17°. 

The significant part of the difference is considered as a 
whole number, or the real difference is multiplied by 10 7 , for 
the same reason as that assigned in App. 1. 24. for the con- 
struction of the tables of proportional parts which are attached 
to the logarithms of numbers. 

S. Ex. l. To find L sin 17°, 14', 12". 

L sin 17°, 14', = 9*4716785 
Now Diff. for l" = 67 '850 
.-. Diff. for 12" = 814-200 = 814-20 



.*. L sin 17°, 14', 12" = 9*4717599. 
Ex. 2. If L sin A = 9*4685537, required A. 
L sin A = 9*4685537 
Z> sin 17°, 6' = 9*4684069 



Diff. = 1468 

Now Diff. for l" is in this case 68*400, and = 21*46; 

M r 68-400 

.*. A = I7°,6',2l"*46. 



167 
Ex. 3. If L cos A = 9*9784328, required A. 

In this case, because the increase of the angle is attended 
by the decrease of the L cosine, we subtract the given L cosine 
from that in the tables which is next greater than it. 

Now L cos 17°, 54/ = 9*9784519 
LcosA =9*9784328 



Diff. = 191 

Now Diff. for l" in this case is 6*800, 

and = 28*088 = 28*09 nearly ; 

6*800 J 

.-. A = 17°,54',28"*09. 

Ex. 4. Required the L cosine of 72°, 5', S" '. 

By the tables, p. l6l, 

L cos 72°, 5', = 9*4880335 , and Diff. for l" = 65*15 ; 
. DifF.for 8"= 521*2 



.*. L cos 72°, 5', 8" = 9*4879814 



The difference for the seconds being in this case subtracted 
from L cos 72°, 5'. P. 158. Ex. 3. 



It may here be observed, that the difference for additional 
seconds must be added for L sines, L tangents, and L secants, 
Arts. 51, 53, 54; and subtracted for L cosines, Art. 52, 
L cotangents and L cosecants. 



168 



9- To shew that the same columns of Differences for l" 
serve for L sin A and L cosec A, for L cos A and L sec A, 
and for L tan A and L cot A. 

i 
For sin ^4 



cosec A 

.*. L sin A = 20 - L cosec A. Art. 8. 

Similarly, L sin (A + l") = 20 - L cosec (J + l") 

.*. L sin (J + l") - Z, sin A = - {L cosec (J + l") - Z, cosec A \ . 

Hence a column of "differences for l"" is printed between 
the columns of logarithmic sines and cosecants ; serving to the 
former as a column of increments for l", and to the latter as a 
column of decrements for l". 

In like manner it may be shewn that 
L cos (A + l") - L cos A = - {L sec (A + l") - Z, sec A\ 
L tan (J + 1,") - L tan J = - {L cot (J + l") - L cot A}. 

Wherefore the columns of cosines and secants have the 
same differences for l", as also have the tangents and co- 
tangents: and it is to be observed that these differences serve 
as increments to the secants (Art. 53.), and tangents (Art. 54.), 
and decrements to the cosines (Art. 52.) and cotangents. 

10. It has been observed, App. III. 5, that the real 
logarithm of a goniometrical function is obtained by subtract- 
ing 10 from the tabular logarithm ; we shall now proceed 

To establish a general rule for supplying the tens 
when the tabular logarithms of goniometrical functions are 
used. 

Let (cos A) n = a . (sin B) m . (tan C) p be any trigonometrical 
formula adapted to logarithmic computation, App. i. 26; B and 
C being given angles, and A being required. 



169 

Then n.l lo cosA=li a+m J 10 sin B+p . l 10 tan C ; 

.*. n.(l0+l 10 cosA)-n.l0=l l0 a+7n.(l0+l l0 six\B)- t m.l0 

+p . (lO+Z 10 tanC) -p . 10 ; 

.-. rc.Z,cosJ = Z i0 a+m.Zsinj&+^.ZtanC + {^-(m+p)} .10. 

And here n, m, p may be whole or fractional. 

Whence the rule, Add to the second member of the 
equation as many tens as the number of times the tabular 
logarithms of the goniometrical functions have been taken in 
the first member exceeds the number of times they have been 
taken in the second member. 

Ex. 1. (tan A) 5 = cos B . (cos Cf ; 

.'. 5.Z, tan A = LcosB + 2 . Z, cos C + {5 - (l +2)}. 10 
= L cos B + 2 . L cos C + 20. 

Ex. 2. (tan A) 2 (sin B) 6 = | (sec C) 4 , 

2 . Z tan J + 6 . Z sin B = Z 10 a - l l0 b + 4Z sec C + {(2 + 6) - 4}. 10 
= l l0 a - l l0 b + 4Z sec C + 40. 

» 2 (sin 5)2 
Ex.3. (tan^) 2 = — — 



2 * 



(cos cy 

- L tan A = Z 10 2 + - Z sin 5 - - Z cos C 



-HHH)} 



5 2 

= Z 10 2 + - Z sin B - - L cos C - i . 10. 



Had both sides of this equation been raised to the 6 th 
power, the fractional indices would have disappeared and the 
value of tan A would practically have been determined much 
more easily. 

Y 



170 

11. Ex, Required the value of A from the equation 

„, 36-61468 . n ■ , „ 

(cos AY = sin 17°, 9 , 4". 

y 11-9 

2Z, cos ^ = l 10 36-61468 - l 10 11-9 + L sin 17°, 9', 4" + 10 ; 
. Z, cos J = 1 (Z 10 36-61468 - / 10 1 1*9 + L sin 17°, 9', 4" + 10) 
Z 10 36-614 = 1-5636472 

Part for 6 = 71 

Part for 8 = 10 



.-. l 10 36-61468 = V5636553 
Z w ll*9 = 1*0755470 



•4881083 

Z, sin 17°, 9', = 9*4696369 
Part for 4" = 273 



(4 x 68-200) 



Add 10- 



2) 19*9577725 

9*9788862 

The next higher L cosine in the tables is that of 17°, 43', 
which is 

9-9788983 

Subtract 9-9788862 



Diff. = 121 

121 

Now Diff. for l" in this case is 6-733, and = 17'97; 

M '' 6-733 

.-. A = 17°, 43', 17" '97. 



171 

12. We shall, lastly, explain the methods used to 
determine small angles from their logarithmic functions and 
conversely. 

When an angle receives a small increment, the Differential 
Calculus enables us to determine without difficulty the con- 
sequent increase of the goniometric function. 

Required the increase of L sin arising from receiving 
a small increment SO. 

By Taylor's Theorem, 

L sin (0 + S0) = L sin + d 9 L sin 9.80 + d\L sin . — — + ... 

Now L sin = 10 + l 1Q sin ; 

1 COS0 1 

.-. dnLsmO = .— — -=- .cot0, 

ti l e \0 sm0 Z 6 10 

dl L sin = (cosec OY ; 

.-. Zsin(0 + ^) -Lsin^ 



13. Now we know that if be very small cosec is very 
large, and, unless SO be not exceedingly small, the second 

{1 (S 6Y} 
(cosec 0) 2 > is of such magnitude that 

it cannot be neglected in comparison with the first difference 

cot 0. SO. In this case therefore, since the increment of 

1 £ 10 5 

Zsin0 does not vary as SO the increment of 0, the quantity 

to be added to L sin for a small increment of cannot be 

obtained by a simple proportion, but must be determined 



172 

approximately by the tedious process of computing the first 
two terms of the series which gives the value of the increment 
of L sin 0. 

There are three methods of escaping this inconvenience. 

14. The first is to construct tables for the first few de- 
grees to intervals of a second, instead of to intervals of a minute 
as the tables in ordinary use are constructed. By this means 
SO must be an angle less than one second and L sin (0 -+- SO) 
may be roughly computed to decimal parts of a second by neg- 
lecting the second term of series (A), — in which case 

L sin (0 + SO) - L sin = ~ cot 0.S0; 
' l e 10 

.-. ALsinOocSO. 



15. Second method. By the following formulae which 
are given by Maskelyne in his Introduction to Taylor's Loga- 
rithms, we can, with the aid of tables of L sines and L cosines 
calculated to every second for a few degrees, determine a small 
angle very accurately to decimal parts of a second from its 
L sine, and conversely. 

03 Q3 

sin = h . .. = nearly. 

2.3 2.3 J 



cos 



0* 2 
= i 1- ... =i nearly. 

1.2 2 J 



sin0_ _^_ / 0'V 
~0~ = 2 ~2~3 



fcr\ 3 i 

1 - — ) nearly, = (cos0)'. 



Let be an angle containing n\ 



then n = - — jj or = n . sin l" ; 
sin 1 

/. sin - n sin l" (cos 0)% 



173 



Writing n" for 0, and taking the logarithms, 
L sin n" = l lo n + L sin l" + ^ L cos 0-1.10; 

.*. Lsinw" = l u n + Lsinl"-!. (10 - LcosO) (i.) 3 

l w n = L sinrc" + -^(10 — Lcos0) - Z sin l"...(ii.) 

Def. The quantity 10 - L cosn" is called the arithmetic 
complement of L cos n" . 

16. It is to be remarked that in using these formulas to 
determine L smn" when n is given, or conversely, we may on 
the second member of the equations take from the tables an ap- 
proximate value of L cos n" without affecting the result, — be- 
cause the variation of L cos n" is exceedingly small when n is 
small as may easily be demonstrated*. The whole matter 
will be rendered more clear by an example. 

Ex. If L sinn" = 7*3217783, required n. 

Now Taylor's tables give, 



L sin 7', 12' 
L sin 7', 13' 



7*3210583 
7*3220624 



L cos 7', 12' 



L cos 7', 12" 



9W99990 
9W99990 



Wherefore the angle is 7', 12" nearly ; and 7'? 12" is the 
approximate value of the angle we must take in the second 
member of (ii.) 



,0 = 1- 



1.2 1.2.3.4 



1- — nearly-, 



1 S Q 2 /0 2 \ 2 I 
.-. JiocosB, or L cos 0-10, = _ "^iq J"2 +M2") + •••$$ 



^ 



.: tf0(Lcose-lO) = - — (0+2- + ...), 
a very small quantity, if be an angle of only a few seconds, 



174 

Now by (ii.), 

Lsinw" = 7*3217783 

i (10 - L cos 7', 12")} ™™™„ 
3 v ' I = -0000003 

or |x -0000010, J 



7*3217786 
L sin l" = 4-6855749 



.-. l lo n = 2-6362037 = Z 10 432*723 ; 
.-. rc = 432-723, or the angle required is f, 12"'723. 

17. In like manner a formula may be established for 
finding L tan n" from n, and conversely. 

sin0 2.3 

Tan = = — nearly ; 

cos 6 2 

2 
tan0 1_ 2.3 V 1 2/ / GT 



2 2 
l l 

2 2 



= fl-~-j = (cos0)-*; 



wsin 1 



(cosw") 3 ; 



2 2 

.*. L tan w" = Z 10 w + L sin l" L cos w" + - . 10 ; 

3 3 

2 
. \ Z tan »" = l 10 n + L sin l" 4- - (10 - L cos rc") (iii.) 



2 
And / 10 rc = Llaxvn" - Lain \" - - (10 -Z,cosr/')...(iv.) 

3 



175 

18. Delambre^s Tables. The third method is to con- 

struct tables up to a degree, which give l 10 — — , or tables 



which give (li — -z \- L sin l"), for angles differing by one 



second. Such tables are, we believe, printed in no collection 
except those of Callet; they may, however, be easily con- 
structed for use in the following manner : 

Let be an angle of n seconds; .*. = nsin\" . 

rm s i n 7 sm n " • " // * 

Then l 10 — - — = l 10 : — T , = l l0 sin n — l lo n - l l0 sm 1 

n.sm I 



L sin ri' - l m n — L sin 1 



// 



7 sin0 _ . „ _.. „ 
•'• *io ~~^ — I- L sin 1 = jL sin w - Z 10 w. 

Similarly, if be an angle of n minutes, we have 

• 1 1 — r— + L sin l f = L sin n - l 10 n. 


19. To determine the sine of a given small angle, or 
conversely, from Delaware's Tables. 

Since sin = — — - . = .wsml ; 



.-. L sin = (7 10 — — + L sin l") + l u n ; 

, . „ , _ sin 

v. Lsinw" = (Z 10 -— - 4- Lsml ) + l 10 n (i.) 



And l lo n = Z, sin w" - (l 10 — — + Z, sin l")...(ii.) 



The most convenient tables are evidently those which 
gi ye ( 1 10— n~ + £ sin l" ) for every second. 



176 
20. Ex. 1. To determine L sin n" by Delambre's Tables. 

Since, as shewn below*, Z 10 - • + L sin l" increases very 



slowly, we may take for l l0 — — \-L sin l", the quantity given 

u 

in the tables for the angle containing that number of seconds 
which is the nearest integer to the given number (ri), and 
by adding l l0 n to this quantity we obtain Lsmn". 

Thus ; If n = 546*25, required L sine 546"*25. 

By Taylor's Tables. L sin 546", or L sin 9', 6", is 7*4227670 ; 
and Z 10 546 = 27371926. 

Therefore, when is an angle of 546", 

Z 10 \- L sin l" = L sin w" - l 10 n Art. 18. 

= 7*4227670 - 27371926 
= 4-6855744, 

the quantity corresponding to the angle 546" which would 

sin . „ 

be given in the tables of Z 10 -— \- L sin 1 . 

u 

.'. By (i.) Art. 19, L sin 546"«25 = 4-6855744+Z 10 546-25 

= 4*6855744 + 27373914 
= 7-4229658. 



„ sine . e 2 e 4 _ e 2 . 

For — =1 -03 + T^O75-" = 1 -6 nearly; 



1 c6 2 ■ /6 2 \ 2 > , . sine 1 /e 26 3 \ 

uo? s+i (?) +■•• 5 ' ••• **• — — Oo(r + "F+--). 

a very small quantity, if d be an angle of only a few minutes. 



sin 6 



177 

Ex. 2. To determine n where L sin n" is given. 

By referring to the tables of L sines in common use we 
see what integral number of seconds is contained in the angle 
whose L sine is next below the proposed quantity : let this 
number be m. Then substituting in (ii.) Art. 19, the quan- 
tity given for l l0 +Z,sinl" corresponding to the angle m", 


we obtain a near approximation to the value of n. 

Thus, If L sinn" = 7*4230612, required n. 
Referring to Taylor's Tables, we find 
L sin 546" = 7*4227670, 
L sin 547" = 7*4235617 ; 
.'. the value of m in this case is 546. 

Now, as we have seen in the last example, the value of 
+ L sin l" for an angle 546" is 4*6855744. 



10 Q 



Therefore by (ii.), Art. 19, 

l lo n = 7*4230612 - 4*6855744 
= 2-7374868 = / 10 546'37 ; 
.-. 546"°37 is the angle sought. 



THE 



ELEMENTS 



OF 



SPHERICAL TRIGONOMETRY 



BY 



J. C. SNOWBALL, M.A. 



FELLOW OF St. JOHN'S COLLEGE, CAMBRIDGE, 



CAMBRIDGE: 

RINTED AT THE PITT PRESS, BY JOHN SMITH, 
PRINTER TO THE UNIVERSITY. 

FOR T. STEVENSON, CAMBRIDGE: 
AND WHITTAKER & CO., LONDON. 



M.DCCC.XXXIV. 



SPHERICAL TRIGONOMETRY. 



INDEX 



CHAPTER I. 

ON CERTAIN PROPERTIES OF SPHERICAL TRIANGLES. 

ART. PAGE 

1. Def. A Sphere 1 

2. Every section of a sphere is a circle 1 

3. Def. Spherical Trigonometry 2 

7. Def. Poles, and Axis of a circle 3 

8. The Pole of a circle is equidistant from every point in it 3 

9 — 10. Methods of finding the Pole of a circle 4 

11. The arc joining the Poles of two great circles subtends the 

angle of inclination of the circles 4 

12. Two great circles are bisected by their intersections 5 

1 3. A side, and also an angle, of a spherical triangle is less than 180° 5 

14. r^r- = sin Pa. P being the pole of AB, and of a b which 

arc AB s v 

lies between the same secondaries 6 

1 5 — 1 8. The Polar, or Supplemental, triangle 7 

19- In the triangle ABC whose angles are A, B, C and sides are 

a,b,c; a +b > c, and « + 6 + c<2-7r;&c 9 

20. A + B + C is > w, and < 3tt 11 

21 — 24. If a — b, then A = B; and conversely 12 

27. Recapitulation of certain properties of spherical triangles 14 



IV INDEX. 



CHAPTER II. 

FORMULAE CONNECTING THE SIDES AND ANGLES OF A SPHERICAL 

TRIANGLE. 

ART. PAGE 

28. Sin A : sin B : sin C :: sin a : sin b : sin c 15 

™ ^ a cos a — cos 6 . cos c _ 

29. COS A = ;— j- — -. 16 

sin 6 . sin c 

30. Cos A . sin C = cos a . sin 6 — cos C. sin a . cos 6 ; &c 17 

31. The same proved independently , 18 

32 — 33. Gauss' Theorem 19 

34. Napier's Analogies, proved from Gauss' Theorem 22 

A A A A 

35. Cos — , sin — , tan — , sin — in terms of a, b, c 23 

36. Cos — , sin—, tan—, sin— in terms of A, B, C 24 

2 2 2 2 

37' Napier's Analogies independently proved 25 



CHAPTER III. 

ON THE SOLUTION OF A RIGHT-ANGLED TRIANGLE. 

38. Napier's rules 27 

32. The triangle ABC having a right angle at C, A and a are of 

the same affection 30 

40. Case in which a right-angled triangle cannot be solved from 

two given parts 30 

42. On the selection of formulae for solving proposed triangles. ... 31 

33. The solution of Quadrantal Triangles 33 

CHAPTER IV. 

ON THE SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 

44 — 53. Solutions of the several cases"of oblique-angled triangles 

from three given parts 34 

54 — 57. Cases in which the solution is ambiguous 39 

58, 59. The radius of the small circle circumscribing a given 
triangle in terms of the angles, and of the sides, of the 
triangle 44 

60, 61. The radius of the inscribed circle in terras of the sides, 

and of the angles, of the triangle 46 



INDEX. 



CHAPTER V. 

ON THE AREAS OF SPHERICAL TRIANGLES, AND THE SOLUTION OF 

TRIANGLES WHOSE SIDES ARE SMALL COMPARED WITH 

THE RADIUS OF THE SPHERE. 

ART. PAGE 

63. Def. ALune 48 

64. The Area of a Lune = .lirr 2 48 

180 

65. The Area of a Spherical triangle = — *— . tt r 2 49 

180 

66. Cagnoli's Theorem 50 

67- Lhuillier's Theorem 51 

68. Other values of E, the Spherical Excess . 52 

69' The solution of triangles whose sides are small compared with 

the radius of the sphere 53 

70. The area of a small triangle being approximately known, to 

find the number of seconds in its spherical excess 53 

71. Example 54 

72. To free the observed angles of a small spherical triangle from 

Errors of observation 56 

73. A series of small triangles will be determined with the greatest 

accuracy if each of the triangles be nearly equilateral 57 

74. Observations on the reasoning employed in the last Article ... 59 
76. Legendre's Theorem for solving a small Spherical Triangle as 

a Plane triangle 5Q 

77 — 79. The Solution of the Chordal Triangle 64 

80. Another method of solving small triangles 65 



The Student is recommended to omit at the first reading. 
Articles 59 — 62. 66 — 79- 



SPHERICAL TRIGONOMETRY 



CHAPTER I. 

ON CERTAIN PROPERTIES OF SPHERICAL TRIANGLES. 



1. Def. A Sphere is a solid bounded by a surface of 
which every point is equally distant from a fixed point, called 
the center of the sphere. 

Def. The line joining the center with any point in the 
surface is called the radius of the sphere. 

2» Every section of a sphere made by a plane is a 
circle. Fig. 1. 

Let ABCD be a sphere of which the center is O ; AFCG 
the curve in which a plane cutting the sphere intersects its 
surface ; OE a perpendicular from O upon the cutting plane. 

Join E with F any point in AFCG, and join FO. Then 
since OE is perpendicular to the cutting plane, it is perpen- 
dicular to EF, a line in that plane, 

.-. EF=\/(OF 2 - OE 2 ), a constant quantity. 

Now £ is a fixed point in the cutting plane, and F is any 
point in the curve AFC. Therefore AFC is a circle whose 
center is E and radius is EF. 

Cor. If the cutting plane pass through the center of the 
sphere, EO vanishes, and EF becomes equal to OF, the radius 
of the sphere. 

* A 



3. Def. The circle in which a sphere is cut by a plane 
is called a great or a small circle according as the cutting 
plane passes or does not pass through the center of the sphere. 

Unless the contrary be expressly mentioned, when hereafter 
we speak of an arc of a sphere, an arc of a great circle is 
meant. 

4. Def. Spherical Trigonometry investigates the 
relations subsisting between the angles of the plane faces which 
form a solid angle, and the angles at which the plane faces 
are inclined to one another. 

For the sake of convenience the angular point is made 
the center of a sphere, which being cut by the plane faces 
containing the solid angle, presents on its surface a figure whose 
sides are arcs of great circles. 

Let ABC, Fig. 2., be a triangle of this kind, whose sides 
AB 9 BC, CA are formed by the intersection of the planes 
AOB, BOC, CO A with the surface of a sphere of which 

O is the center. The angle of any face, as AOB, is — — — , 

PL Trig. 84 ; and the angle contained between any two faces, 
as BAO and CAO, is the angle contained between AD 
and AE, which are drawn in the planes BAO and CAO at 
right angles to their intersection AO, Eucl. xi. Def. 6., and 
therefore touch the arcs AB and AC respectively. 

5. The angle contained between the planes BAO and 
CAO which intersect in OA, we shall hereafter indicate by 
BAC, or by the letter A ; and the angle BOC subtended 
at the center of the sphere by the side BC which is opposite 
to the angle A, will be represented by a. In like manner the 
other parts of the triangle will be indicated. 

6. After certain properties of spherical triangles have 
been proved, it will not, in pursuing further investigations, 
be requisite to represent in our figures the center of the sphere 
on which the triangles are described. When the center of 



the sphere is not represented we shall often hereafter say " the 
angle AB", — or simply " AB" ; by which we mean the angle 
which AB subtends at the center of the sphere; that is, the 

angle — , or the number of degrees which A OB con- 
radius 

tains, according as the magnitude of the angle is estimated by 

one measure or the other. PL Trig. Arts. 88. 8. 

7- Def. If OE, Fig. 1., be produced both ways to meet 
the surface of the sphere in B and D, these points are re- 
spectively called the nearer and the more remote poles of the 
circle AFC, and the line BOD is called the axis of the circle 
AFC. 

8. The pole of a circle is equally distant from every 
point in it. Fig. l. 

Join B with F any point in AFCG. Then 3 BEF being 
a right angle, 

BF 2 = BE 2 + EF 2 

= BE 2 + OF 2 - OE\ a constant quantity. 

And F being any point in AFCG, B is equally distant from 
every point in that circle. 

Similarly, DF 2 = BE 2 + EF 2 

= DE 2 + OF 2 - OE 2 , a constant quantity. 

Hence D is equally distant from every point in AFCG. 

9. Since BO is at right angles to the plane AFC, every 
plane through BO is at right angles to that plane. Hence 
the angle between any circle whatever and a great circle 
passing through its pole is a right angle. 

If AFCG become a great circle, OE vanishes, E coincides 
with 0, and BOF becomes a right angle. See Fig. 3. Where- 
fore BFA, the angle of inclination of the planes BOF and 
AOF, is a right angle ; and since BO is perpendicular to 
the plane AOF, and therefore to OF a line in that plane, 



4 



BOF is a right angle, and BF is a quadrant, or the fourth part 
of the circumference of the great circle BFD. Hence it 
appears, that the extremity J? of a quadrant FB drawn at 
right angles to a great circle AFC is the pole of that circle. 

Cor. If B be the pole of AFC, AO and FO are at right 
angles to BO, and AOF is the inclination of JOB and FOB ; 

.'. Art. 6., LAF= lABF. 

Def. Great circles which pass through the pole of a 
great circle are called secondaries to that circle. 

10. If from a point on the surface of a sphere there can 
be drawn two arcs {which are not parts of the same circle) 
at right angles to a given circle, that point is the pole of 
the circle. Fig. 3. 

Let B be the point, BA and BF arcs through A and F, 
points in the circle AFC; and let BA and BF be at right 
angles to AF. 

Then since BA and BF are at right angles to AF, and 
are not parts of the same circle, their planes BAO and BFO 
intersect, and their intersection BO is at right angles to 
AOF, Eucl. xi. 19; — therefore B is the pole of AFC. 

Cor. Hence also it appears, that the intersection of two 
arcs drawn at right angles to a given circle through any two 
points in it, is the pole of that circle. 

11. The angle subtended at the center of the sphere 
by the arc which joins the poles of two great circles, is the 
angle of inclination of the planes of the circles. 



Fig. 4. 



Let the given circles be FD and FE intersecting in F, 
A and B their respective poles, and ABD the circle through 
A and B. 

Now AO is perpendicular to OF, a line in the plane DOF 
And BO is perpendicular to OF, a line in the plane EOF i 



.*. OF is perpendicular to the plane AOB, and therefore 
to OD and OE, lines in that plane; 

.-. DOE is the angle of inclination of FOB and FOE. 
And AOB = JOB - BOD = 90° - BOD 

= BOE - BOD 

= D0E. 

Coe. Hence also it appears that z AB - Z DE = l DFE> 
Arts. 6. 5. 

12. Two great circles are bisected by their intersections. 
Fig. 3. 

Let BAD and BED be parts of two great circles intersect- 
ing each other in B and D. Join BD, and let O be the 
center of the sphere. 

Then since BAD is a great circle, O is a point in its plane. 
Similarly, O is a point in the plane of BFD ; therefore O lies 
in the intersection of these planes, that is, in the line joining B 
and D. Therefore BAD and BFD are semicircles, having 
BOD for a common diameter, and the two great circles are 
bisected by the points B and D. 

13. Any side of a spherical triangle is less than a semi- 
circle, and any angle is less than two right angles. 

Since Euclid takes two right angles as the limit of a plane 
angle, this is also his limit for the angle of any plane face 
of a solid angle*. Hence in a spherical triangle no side can be 
equal to a semicircle. 

* If a face could have an angle greater than two right angles, Euclid's Pro- 
position, xi. 20., " If a solid angle be contained by three plane angles, any two of 
them are together greater than the third," would not be true in all cases. 

For let ACB, Fig. 6., be a circle whose diameter is AOB, and let COA revolve 
about OC, so that A comes into the position a. Now if AOa be less than BOC, 
it is evident that if the faces of the solid angle whose edges are AO, aO, CO be 
AaO, a CO, AECO, two of these, viz. A a and a CO, are together not equal to the 
third AECO. 

If, following Euclid's construction in xi. 20., we made AOE = AOC, and joined 
AE. Euclid's proof would fail from EA not cutting the edge OC. 

The faces of the solid angle whose edges are AO, a 0, CO, are ADCO, CaO, aAO, 



6 

Thus if ACS, Fig. 5, intersect AEBF in the points A and 
B, and CD be any other arc, then the triangle connecting the 
points B, C, D is not the figure formed by BC, CD, DAFB, 
(of which the side DAFB is greater than a semicircle), but the 
figure formed by BC, CD, BED. 

If a side, as ADB, become a semicircle, then the arcs which 
join A and B with any other point C, are portions of the semi- 
circle BCA, and the arcs joining the points A, B, C cease to 
form a triangle. Wherefore the side of a triangle is less than 
a semicircle. 

Hence it follows that an angle of a triangle must be less 
than two right angles. 

For if possible let BFADC, Fig. 5., be a triangle having 
BCD greater than two right angles. Produce BC to A ; then 
BFA is a semicircle, and BFAD is greater than a semicircle ; 
which is impossible. 

If BCD become equal to two right angles, BC and CD 
become parts of the same circle, and the figure ceases to be a 
triangle. 

Wherefore the angle of a triangle is less than two right 
angles. 

Cor. Hence it follows, that if the great circle be completed 
of which any side AB of a triangle ABC forms a part, the 
triangle lies within it. 

For if possible, let C, Fig. 7., lie without the circle, and let 
CA cut it in D\ then since DA is a semicircle, Art. 13., 
CD A is greater than a semicircle; which is impossible. 

14. If P be the pole of a great circle BAC and of a small 

circle bac which are cut by the great circles Pa A and PbB, 

arc ab . 

then — — - sin V a. rio-. s. 

arc AB 6 



Let OP, which is perpendicular to the planes of both 
circles, cut the plane of the small circle in D. Join AO, BO, 
aD,bD, aO. 

Then aD is perpendicular to PO, and therefore parallel 
to AO. 

Similarly, bD is parallel to BO ; 
.-. L aDb = z AOB. Eucl. xi. 10. 
arc ab arc AB 



that is, 



aD AO 

arc ab aD aD 



arc JB AO aO 



sin a OP, or sin Pa, 



15. Def. If the angular points A, B, C of a spherical 
triangle ABC be made the poles of the three great circles 
DE, EF, FD respectively, the triangle ABC is called with 
respect to DEF the primitive triangle, and DEF is called with 
respect to ABC the polar triangle. Fig. 9. 

16. The angular points of the polar triangle are the 
poles of the sides of the primitive triangle. Fig. 9. 

Join AE and BE, by arcs of two great circles. 



Then, since A is the pole of ED, AE = — , 



7T 

And, since B is the pole of EF, BE = — . 



Therefore the great circle of which E is the pole passes through 
A and B, or E is the pole of AB. Similarly D and F are the 
poles of AC and BC respectively. 



8 



17- The polar triangle is called also the supplemental 
triangle, from the following property it possesses. 

The sides and angles of the polar triangle are the supple- 
ments of the angles and sides respectively of the primitive 
triangle. 

For zi= HG. Art. 10. Cor. 
= EG - EH 
= EG- {ED - DH) 
=EG+DH- ED. 
But EG and DH are quadrants, or each subtends an 
angle ~; 

.-. z 4 = TV - ED. 



Similarly, z B = tt - EF, and z C = tt - F2X 
Again, AB = AG - BG = AG - (IG - BI) 
= AG+BI-IG 
Since E = /G. Art. 10. Cor. 



7T 



+ --2Z. 

= w-E. 
Similarly, BC = tt - F, and JC = tt - Z>. 

18. If a general equation be established between the sides 
and the angles of a spherical triangle, a true result is obtained 
if in the equation the supplements of the sides and of the angles 
respectively be written for the angles and sides which enter 
into the equation. 

For if the equation be proved for any triangle whatever, 
we may substitute in it the sides a, &', c and the angles 
A\ B', C of the polar triangle in the place of the sides a, b, c 
and the angles J, B, C of the primitive triangle. And in the 
equation as it thus stands, putting for the sides and angles 
of the polar triangle their equivalents drawn from the primitive 
triangle, viz. 



7r - A = a, 7t - a = A, 

7r-B=b\ w-b = B\ 

7T - C = c', 7T - C = C, 



we obtain a true result, which differs from the original equa- 
tion, in having the supplements of the sides and of the angles 
respectively written for the angles aud sides of the triangle. 

Ex. If A, B, C be the angles, and a, 6, c be the sides of 
any triangle, it will hereafter be shewn that, 

cos a — cos b . cos c 

COS A = ; ; . 

sin 6 . sin c , 

r¥ _. ., cos a' — cos b' . cos c 

1 nerelore cos A : — —. — : — : . 

sin 6 . sin c 

Now A' = 7r - a, a = 7r - A, b' = ir - B, c = it - C ; 

COS (7T - J) - COS (7T - B) . COS (7T - C) 

.'. cos (tt - a) = 2 r~7 ^ — r— 7- 7^ • 

sin (tt - B) . sin (tt - C) 

cos A + cos Z? . cos C 

.*. COS a = ; r— :; . 

sin 5 . sin C 

19- Any two sides of a spherical triangle are together 
greater than the third, and the sum of the arcs which are the 
sides is less than the circumference of a great circle. Fig. 2. 

For, Eucl. xi. 20., any twb- of the angles AOB, BOC, 



which form the solid 



^ _ . / arc AB arc BC arc CA\ 

CO A* or , , — -— — , 

V AO ' AO ' AO J 9 

angle at O, are together greater than the third. Therefore any 

two of the arcs AB, BC, CA are together greater than the 

third. 

* B 



10 

Also, Eucl. xi. 21., the three angles forming the solid angle 
at are less than four right angles ; 

arc AB arc BC arc CA 



.-. arc AB + arc BC + arc CA < 2-rr . A0 9 i. e. the 
circumference of a circle whose radius is AO. 

Cor. 1. Since the sum of the plane angles which contain 
any solid angle is less than four right angles, Eucl. xi. 21., it 
follows from the same mode of proof, that the sum of the sides 
of any polygonal figure described on a sphere is less than the 
circumference of a great circle. 

The polygonal figure, however, must be such that its 
area is contained on the surface of a hemisphere. For in the 
proof of the proposition of Euclid to which we have referred, 
a plane is supposed to cut all the plane faces which contain the 
solid angle, — which cannot be the case unless all the edges of 
the solid angle lie within the same hemisphere. 

Cor. 2. Also let ABCDE be a five-sided figure described 
on a sphere, and let it be divided into triangles by the arcs 
AC, AD. (The Student will easily draw the figure.) 

Then AB + BC>AC; 

.-. AB + BC + CD>AC+CD, and AC+CD> AD; 

.-. AB+ BC+CD>AD; 

.*. AB+ BC + CD+ DE > AD + DE> AE. 

And the same method of proof being applicable to a polygon 
of any number of sides, it follows that the sum of all the sides 
save one of a spherical polygon is greater than the remaining 
side. 



11 



Cor. 3. If a and b be two sides of a spherical triangle, 
since each is less than 7r, 

.-. a + b < 2 tt, .*• i (a + 6) < 7t. 



And a — 6 < 7r, 



••• i(a-6)< 



Also since an angle of a triangle is less than 7r ; 
.-. A + B < 2tt, .*. ±(A + B) < it, 



And A - B < w, 



W-*)<=;- 



Cor. 4. a', &', c', being the sides of the polar triangle, 

a' + 6' > V ; 

... („. _ j) + („. _ 5) > (tt - C) ; 

.'. 7T>J +B-C 

Or, ^ + 5 - C < tt. 

20. The sum of the angles of a spherical triangle is 
greater than two, and less than six, right angles. 

Let A, B, C be the angles, and a, 6, c the sides of a 
triangle; A', B', C and a', b', c the angles and sides of its 
polar triangle. 

Now 27r>a + b' + c. Art. 19. 

> (tt - A) + (tt - B) + (tt - C) 
>3tt- (A + B + C); 
.-. A + B+ C>tt. 
Again, since each of the angles A, B, C, is less than 7r, 
.-. A + B + C<3tt. 



12 



21. The angles at the base of an isosceles triangle are 
equal to each other. Fig. 10. 

Let ABC be an isosceles triangle, having AB = AC, and 
therefore z AOB = z AOC 

From D any point in OA draw DG perpendicular to the 
plane BOC, and therefore at right angles to every line it meets 
in that plane ; and in that plane draw GE and GF perpendi- 
cular to OB and OC ; join ED, FD, OG. 

Then OE 2 = OG 2 - GE 2 

= (0D 2 - DG 2 ) - (ED 2 - DG 2 ) 
= OD 2 - ED 2 ; 

.*. DE is perpendicular to OE, and .'. Z DEG= inclination of 
the planes BOC and BOA = £ B. 

Similarly DF is perpendicular to OC, and z DFG = lC. 
Now DE = 0D. sin J 05 = 0Z> . sin JOC = DF. 
And J£G 2 = DE 2 - DG 2 = Z>F 2 - DG 2 = FG\ 
Hence, since GE, ED = GF, FD, and (2D is common, 
.-. z ZXEG = z DFG, orz5=zC. 

22. Conversely, if z 5 = z C, or z i)£G = Z DFG, it 
may be shewn from the same figure that AB = AC ; — that 
is, the angles at the base being equal, the sides opposite to 
them are equal. 

Cor. Hence also it follows that every equilateral tri- 
angle is also equiangular; and conversely, that every equi- 
angular triangle is also equilateral. 

23. Of the two sides which are opposite to two unequal 
angles in a. triangle, that is the greater which is opposite to 
the angle which is the greater. Fig. 11. 



13 

Let z ABC be greater than z BAC Then AC > BC. 

Make z DBA = z ZM# ; .•; DA = DB. 
But CD + DB>BC. 

And AC=CD + DA=CD+DB; 

24. Conversely, it may easily be shewn that of two angles 
in a triangle opposite to unequal sides, that is the greater which 
is opposite to the side which is the greater. 

Cor. Hence A ~ B and a — b have the same sign. 

25. Article 23 has been proved after the manner of 
Euclid i. 19. The following propositions may be enunciated for 
spherical triangles, in the terms used for plane triangles, and 
may be proved in nearly the same words. Euclid 1. Props. 4. 
8. 24. 25. 26. 

26. If a semicircle pass through a point on a sphere and 
the pole of a given circle, (and is therefore perpendicular to 
this circle,) of all arcs which can be drawn from the point to 
the circle, the least is the shorter segment of the semicircle, 
and the longer segment of the semicircle is the greatest ; and 
an arc nearer to the least is less than one more remote. 
Also, there can be drawn but two equal arcs from the point 
to the given circle, one on each side of the shortest arc. 
Fig. 12. 

Let DBC be the given circle, P its pole, BAPb the semi- 
circle through P and the given point A ; PO is at right angles 
to the plane of DBC, In the plane bPAB draw AM parallel 
to OP ; it is therefore at right angles to the plane of DBC. 

Take BOD any angle, and make Z BOC = Z BOD, 

Now since M is a point in the diameter bOB of the circle 
b DB, MB is the least line that can be drawn from M to the 
circle, and Mb is the greatest. 



14 

And AD 2 = AM 2 + MD 2 > AM 2 + MB 2 > AB 2 . 

Since AO, BO=AO, OD, but DA > AB ; 

.-. z AOD > z AOB, or arc AD > arc AB. 

Again, AD 2 = AM 2 + MD 2 < AM 2 + Mb 2 < A b\ 

Whence as before, arc AD < arc APb. 

And since AD is any arc, APb must be the greatest arc 
that can be drawn from A to the circle BDb, and AB is the 

Again, if AE be any other arc cutting bDB in 22 a point 
more remote from B than Z> ; 

JZ> 2 = AM 2 + MD 2 < AM 2 + Jf£ 2 < J£ 2 . 
Whence, arc JZ> < arc vIjE. 

Also if jBOC be made equal to BOD, it may easily be 
shewn that MC = MD, and arc AC = arc J.Z). And there 
can no other arc equal to AD be drawn from A to DC; 
otherwise the arc more near to AB would be equal to the one 
more remote, which has been shewn not to be the case. 

27- Before we proceed to prove certain formulae which are 
useful for the solution of triangles, we will recapitulate some 
properties of triangles to which we shall have occasion to refer. 

1. A side must be less than a semicircle. 

2. An angle must be less than two right angles. 

3. Any two sides are together greater than the third. 

4. The greater side is opposite to the greater angle ; and 

conversely. 

5. A - B and a — b are of the same sign. 

6. a + b + c <2<7r. 

7- A + B + C > 7T, and < Sir. 

8. « + 6<27r, a-b<7r. 

9- A + B<2ir, A - B < 7T. 

10. A + B -<7<7T. 



CHAPTER II. 

FORMULA CONNECTING THE SIDES AND ANGLES OF A SPHERICAL 

TRIANGLE. 



28. The sines of the angles of a triangle are proportional 
to the sines of the angles which the sides respectively opposite 
to them subtend at the center of the sphere : Or, Art. 6. ; The 
sines of the angles of a triangle, are proportional to the sines 
of the sides respectively opposite, that is, 

sin A : sin B : sin C :: sin BC : sin AC : sin AB. Fig. 10. 

Let ABC be the triangle, O the center of the sphere, D any 
point in OA. Draw DG perpendicular to the plane BOC, DE 
and DF in the planes AOB and AOC perpendicular to OB 
and OC. Join EG, FG. 

Then, since DG is perpendicular to the plane BOC, and 
therefore to EG, a line in that plane, 

EG 2 = ED 2 - DG 2 = (OD 2 - OE 2 ) - (OD" - OG 2 ) = OG 2 - OE 2 ; 

.'. EG is perpendicular to OE. 

And ED, EG being each perpendicular to OB, the inter- 
section of the planes A OB and BOC, the angle DEG is the 
inclination ( z B) of those planes. 

Similarly, L DFG = L C 

Now DE . sin B = DG = DF . sin C ; 

DF 
sinB _DF_OD _ sin AOC _ sin AC 
J '' sin C ~ DE~DE~ sin A OB = sin AB % 
OD 



16 

And by letting fall a perpendicular from any point in OB upon 
the plane AOC, it might in like manner be shewn that 

sin A sin BC 
sin C sin AB 

Wherefore, 

sin A : sin B : sin C :: sin BC : sin AC : sin JZ? (i.) 

In the figure we have supposed the perpendicular from D 
on the opposite plane to fall within the angle BOC, — if it fall 
without it, the proposition admits of a proof in the same 
words. 

29. To express the cosine of an angle of a triangle in 
terms of the cosines and sines of the sides. Fig. 13. 

From any point D in OA draw DE and DF in the planes 
AOB and A OC at right angles to AO. Therefore Z EDF is 
the inclination of those planes to each other, that is, the angle A. 
Join EF. 

Then from the triangles FOE and FDE, 

OF 2 + OE 2 - 20F. OE . cos FOE = FE 2 

= FD 2 + DE 2 - 2 FD . DE . cos FDE ; 

.-. 2 0F.0E. cos FOE 

m (OF 2 - FD 2 ) + (OE 2 - DE 2 ) +2FD.DE. cos FDE, 

= 2 02> 2 + 2. FD.DE. cos FDE ; 

0Z> 0Z> FD DE 



/. cos FOE, or cos « = — — -. 



+ 



cos FDE 



. cos A 



OF OE OF OE 

= cos b . cos c + sin b . sin c . cos ^f ; 
cos « — cos b . cos c 



sin 6 . sin c 



(n.) 



17 

Cor. By writing ir-a for A, tt-A for a, &c. (Art. 18.), 
we have 

cos A + cos B . cos C 

cosa= —- — —- (in.) 

sin B . sin C 



30. To shew that 

cos A sin c = cos a . sm b — cos C . sm a . cos b. 

_, ,... _, cose - cos a . cos b 

± rom (n), cos C = : : . 

sin a . sin 6 

.*. cos c = cos a . cos & + cos C . sin a. sin b. 

So also, 

cos .4 . sin b . sin c = cos a — cos b . cos c 

=s cos a— cos b . (cos a . cos 6+cos C . sin a . sin b) 

= cos a . { 1 — (cos 6) 2 } -cos C sin a . sin b . cos b ; 

.*. cos J. . sine = cos a . sin 6 - cos C. sin a. cos & (iv.) 

^ «. , • . . sin C 

Cor. 1. Since cos A . sm c = cos ^.smffl. — — - 

sm A 

= cot .4 . sin a . sin C, 
we have from (iv.), 

cot A . sina.sinC = cos a . sin b — cos C . sin a . cos b ; 
.*. cot .4 . sin C = cot a . sin b — cos C . cosb (v.) 

Cor. 2. By writing ir - a for A, &c. Art. 18., we have 
from (iv.) and (v.), 

cqs a . sin C = cos ^f . sin B + cos c . sin A • cos 5 (vi.) 

cot a . sin c = cot ^ . sin 5 + cos c . cos 5 ( vu ^) 

*C 



18 

31. To prove the formula (iv.) by an independent 
method. 

Taking the construction of Art. 28., produce EG to meet 
OC in H. Fig. 10. 

EH EG GH 
Thensma = — = — + _ 

EG GF 
= 0H + ^E' hj ^ 0EH > GFH 

EG ED EO GF FD OD . 



ED EO OH FD OD OE 



n sin c . 1 

= cos B . . cos a + cos C. sin b . 

cos c cos c 



.*. cos C . sin b = cos c . sin a — cos B . sin c . cos a ; 
which is the same as (iv.). 

Cor. Hence (ii.) can be proved without taking for granted 
the Proposition in Plane Trigonometry, 

6 2 + c 2 - a 2 

cos A = . 

26c 

For cos a . sin c . cos B = cos c . sin a — sin 6 . cos C, , 

So cos a . sin b . cos C = cos 6 . sin « - sin c . cos B, 

and substituting this value of sin c . cos B in the first equation, 

cos a . (cos & . sin a— cos a . sin b . cos C) = cos c . sin a— sin& . cos C; 

.*. cos C . { 1 — (cos a) 2 } . sin 6 = cos c . sin a — cos a . cos 6 . sin a ; 

cos c — cos a . cos b 



.-. cos C 



sin a . sin 6 



EG 

* Note. The quantities introduced into the numerator and denominator of {jtj , 

are those which connect EG and OH by means of the sides which are common to 
each two of the triangles DEG, DOE, OEH. Thus by introducing ED we connect 
the triangle DEG, in which EG is, with the triangle DEO; and by introducing EO 
we connect DEO with OEH, the triangle in which OH is. In like manner GF 
and OE are connected through the triangles GFD, FDO, DOE. 



19 

32. The following theorem, was first given in this form 
by Gauss. 

c C 

If, p = cos - . sin \{A + 2?), P = cos — . cos \ (a — b), 



c C 

q = cos - . cos -J- {A + B), Q = sin — . cos \{a + 6), 

c C 

r = sin- . sin \{A — B), R = cos— . sin ±-(a — b), 

At At 

c C 

s = sin-.cos-|-(J - B), S = sin~ . sin -£-(« + 6), 

At At 



Then p . q, p . r, p . s, q . r, q . s, r . s are respec- 
tively equal to P . Q, P . R, P . S, Q . R, Q . S, R . S. 

sin 5 sin b 

0). li-7— 7=i±-r- ; 

sin J sin a 
.". sin a (sin ^ ± sin B) = sin J (sin a ± sin 6) ; 
.". sin c (sin J ± sin 5) = sin C . (sin a ± sin 6) i. 

(2.) By (iv.), cos A . sin c= cos a. sin 6 -cos C. sin a. cos 6. 

So cos JB . sin c=cos b . sin a— cos C . sin 6 . cos a ; 
/. sine. (cosl?± cos ^f) = (l =pcosC) .(sina.cos&±cosa.sin&) 
= (l =f cos C) . sin (« i 6) n. 

(3.) Writing ir - C for c, Sec. Art. 18, and transposing the 
terms of n., we have 

(l ± cose) . sin (A ± B) = sinC. (cos b ± cos a) in. 

Now expressing the forms i, n, in, in terms of the sines and 

C c 
cosines of — , - , \(A ± B), ^-(a ± 6), we have 



20 



<o 



I 

8 



+ 



+ 
8 



+ 

S3 



.£ 

"co 
II 

I 



I 

.s 

*co 



.s 

"co 



+ 



+ 

_c 
"co 

.5 

"co 
C3 



i 



O|o< 



eq 
I 

% 
.£ 

"53 



i*0 
+ 



+ 

8 



O |c* O | o* ^ 



CN 



| 0* 


O|o* 


co 


.2 


II 


II 




£ 


+ 


1 


3 




> 


H» 


o 


CO 

O 
to 



fci &, 



.s 

"co 
to I O* 

CO 

O 
O 

a 



«3 
3 

*CO 



to I o* 

CO 

O 
to 



to I O* 




01 

+ 

• I— ( 
CO 

to I O* 

to 
O 
O 



CO 

O 
to 



I 



to I cn 

G 



to I ©* 

G 



&3 fi$ 






o 



s 
S 






J3 



21 

33. The quantities p, q, r, s are respectively equal to 
P, Q, R, S. 

Since pq . pr = PQ . PR ; 

. p \ qr = P 2 .QR; 

.'. p 2 = P°. 

Similarly, q 2 = Q 2 , r 2 = P 2 , s 2 = tf 2 ; 

,\ p = ± P, g = ± Q, r = ± P, = ± & 

Now in a spherical triangle, Art. 27., 

-^(^4 + 2?) and -^-(a + &), are each less than 7r, 

■J- (-4 - B) and i(a - 6), are each less than — , 
and are of the same sign, 

— and - are each less than — ; 

2 2 2 

.*. p, r, s, P, P, S are all positive quantities. 
Also since pq = PQ, and |? = + P, /.</ = + Q ; 

c C 

.'. p = + P, or cos - . sin-i-(^ + P) = cos— . cos \{a -b)... (viii.), 

c C 

q = + Q ? cos- . cos-J-(^-f-P) = sin— . cos-|-(a + 6)...(ix.) 



c C 

+ P, sin -. sini(^-P) =cos — . sin -§-(«-&) — (x). 

C C 

+ S, sin- . cos^(J-P) = sin — .sin i(a + 6)...(xi.) 



* 



22 



Cor. From (ix.) it appears that cos \ (A + B) and 
cos \- (a + b) are of the same sign, and therefore \ ( A + B) and 

\ (a + b) are both greater or both less than — . 



Def. When two angles are both greater or both less than 
a right angle, they are said to be of the same, or of like affec- 
tion. 

Thus \ ( A + B) and \ (a + b) are of like affection. A pro- 
perty of spherical triangles which may be added to those 
enumerated in Art. 27. 

34. Since 

V P .,j^ cos ±-(a-b) C , .. x 

J - = -; /. tanjr(A + B) = ~ -<-. cot- (xn.), 

r R , , . _. sin-i(«-&) C 

— = -; .'. tan-i-(^-5) = -r-T7 77 • cot - ( xm -)> 

s # 2V ' smj(« + 5) 2 

S s \ cos \{A-B) c 

— = -, .'. tanf(a + o) = — — — — - .tan— (xiv.), 

Q q 2V ' cos jr(A + B) 2 v " 

R r ,, sin jr(A-B) c , x 

— = -, .-. tani(a -b) = . , ; , ' . tan - (xv.). 

P p sin ±(A + B) 2 v ' 

These four equations are called Napier's analogies, and are 
of great use in the solution of triangles*. 



* The quantities p, q, r, s, P, Q, R, S, will be most easily remembered from 
Napier's first and second analogies written thus, 

sin ±(A + B) .cos ^ cos £ (a -6) .cos ^ p 



P_ 

cos l(A + B) .cos J cos l(a + b) . sin ^ 



ry ^ __* __,,.. . LN _ : _C Q 



sin %(A-B) .sin 2 sin £(«-&') .cos ^ jj 
_ - _ = - = - 

cos £(.4 -.B). sin ^ sin \ (a + b) .sin ^ 



23 

35. The following expressions can, without much diffi- 
culty, be proved from the formulas of Art. 32. ; we will how- 
ever establish them from the fundamental formulae (i), (ii), (iii). 

AAA 

To find the values of cos — , sin — , tan — , and sin A 

in terms of a, b, c. 

_ ... . , cos a - cos b. cos c 

By (n.), cos A - 



( cos i) 



sin b . sin c 

A\ 2 . cos a — cos b . cos c 

1 + cos A = 1 + 



sin b . sin c 
.\ 2 ( cos — J . sin b . sin c = cos a — (cos 6 . cos c — sin b . sin c), 

= cos a — cos (6 + c), 
= 2 sin \{a + 6 + c) . sin \{b + c - a) ; 
PL Trig. Art. 44. (4.). 
And if # = iO+&+c); •'. S-a= ^(a + b + c) -a = i(6 + c-«), 
J^ 2 sin # . sin (*? - a) 
sin 6 . sin c 



cos — = 
V 2/ 



J /sin *S . sin (# - a) 



cos — = \/ : — - — ; ( x vi.) 

2 sin b . sin c 

-4\ 2 , cos a — cos 6 . cos c 



Again, 2 ( sin — ) = 1 - cos A — 1 



sin 6 . sin c 



A /sm(S-b).sm(8-c) 
Whence, sin — = \/ — . [ _._ _ ... (xvn.) 



2 v sin b . sin c 

cirj 

A 2 /sin (£ -6). sin OS -c) , 

Also, tan - = = V • c* we \ '" ( xvm ') 

2 J v sm S . sm (S - a) ' 

cos — 

2 

• J J 

And sm A = 2 sin .— . cos — 

2 2 

2 

/y/ { sin S. sin (*S-«). sin (>S-6) . sin (S-c) ] . . . (xix.) 



sin b .sine 

The positive signs of the square roots are taken in these 

cases, because — is necessarily less than — . 



24 



36. To find the values of cos - , sin - , tan - , and sin a 

lit Al A) 



in terms of A, B, C. 

_, ..... , cos J + cos B. cos C 

rJy (in) we have, cos a = 

J v ' sin 5 . sin C 



.-. 2 (cos — ) = 1 + 



cos ^ + cos B . cos C + smB . sin C 

cos # = : : 

sin B . sin C 

cos A + cos (B - C) 
sin 5 . sin C 

_ 2cos±(A + C-B).cos±{A+B-C) 
sin B . sin C 

And if 6" = i(A + B + C) ; then S' - B = ^A + C-B) ; 

a /cos (S' - B) . cos (#' - C) , , 

.-. cos- = V . ' . (xx.) 

2 v sm jB . sin C 



( • «V 
Again, 2 I sm - I = 1 - cos a - 1 



cos A + cos B . cos C 

sin B . sin C 



cos A + cos (J? + C) 
sin jS . sin C 

a / cos tf'. cos OS" - A) , * 

Whence sm - = \/ : — - — : — (xxi.) 

2 v sin B . sm C 

. a 

a 2 / cos S' .cos (S'- A) , 

Also, tan - = = V r <y P v fQf n \ "" V xx11 ') 

2 a v cos(6 -Z?).cos(*y -C) 
cos - 

2 

. a a 

And sm a = 2 sin . cos - 

2 2 

:-T-=^-7^v / |-^s^cos(^-J)xos(y-5).cos(y-C , )J (xxiii). 
smB.smC c 



25 

Note. Since the angles of a spherical triangle are together 
greater than two and less than six right angles, S' is greater 
than one and less than three right angles, and its cosine is 
therefore a negative quantity. Also since, Art. 27-, A + B—C<ir, 
therefore cos ^ (A +B— C), or cos (S f — C), is a positive quantity ; 
and in like manner cos (S' — B) and cos (S' — A) are positive. 
Wherefore the three last formulae are not imaginary quantities, 
as at first sight they appear to be, but real quantities, as they 
ought to be. 

37. Before leaving this part of the subject, we will prove 
Napier's Analogies from the formulae (i.), (ii.), (iii.). 

_ ,.,. . cos A + cos B . cos C 

By (m.), cos a = . . ; 

7 sin B . sin C 

.'. cos A + cos B . cos C = cos a . sin B . sin C. 

So cos B + cos A . cos C = cos b . sin A . sin C 

By addition, (cos A + cos B) (l + cos C) 

= (cos a . sin B + cos 6 . sin A) . sin C; 



.*. (cos A + cos B) . 2 I cos — ) 



, . . sin 6 -.,*.£ C 

(cos a .smA . — h cos .smA) .2 sin — . cos — ; 

sin a 2 2 



sin A . C 

cosA+cosZ*=-; . (cos a . sm 6+ cos 6 . sin a), tan — 

sin a '2 



smA . , 7X C 

— : . sin ( a + o) . tan — ; 

sin a 2 



26 

sin B s 



. . . „ . ■'■ / sm2?\ 

Again, sm A + sin B - sin A . 1 + - 

V sin A) 

I sin b 

= sm J . l + — 

\ sm a 

sin J 

= — . (sm a + sin b) ; 

sin a 

sin A + sin 5 sin « + sin 6 C 

.♦. — _ . — _ . C ot — , 

cos A + cos B sm (a + 6) 2 

2.sin-i-(J + #).cos-i-(J-i?) 2.sini(a+&).cos 1 L-(a-&) C 
2.cosi(J + Z?).cosi(J-i?) ~ 2. sin ^ (a +b). cos ±(a+b)' 2' 

, ^ j ™ cos 4- (a — 6) C . . 

.'. tan±(A + B)=: ±) it- cot- (a). 

v l cos^(a + b) 2 

In like manner, 

sin A — sin B sin a — sin & C 

— — . . cot — ; 

cos A + cos B sin (a + b) 2 

Whence tan \{A - B) = '?" *** " Q .cot- ()3). 

2 v ' sin-i-0 + 6) 2 v/ 

By writing 7r- J for a, &c, Art. 18., these two formulae become 

, v cos\(A-B) c , . 

t an i(a+ 6) %os ^ + jB) .tan- 2 ( 7 ). 

sin^-M-i?) c ,* N 

tan-i-(« - b) = — fy— - —.tan- (S). 

These last two analogies might have been easily proved inde- 
pendently from the formula, 

. cos a — cos b . cos c 

cos A = : — - — : . 

sin o . sm c 



CHAPTER III. 



ON THE SOLUTION OF RIGHT-ANGLED TRIANGLES. 



38. Napier s Rules. Right-angled triangles are solved in 
the following manner. 

The right angle being left out of consideration, the two 
sides including the right angle, and the complements of the 
hypothenuse and of the other angles, are called the circular 
parts of the triangle. Any one of these being fixed upon as 
the middle part (if), the two circular parts immediately join- 
ing it are called the adjacent parts (J i? A 2 ), and the other two 
parts are called the opposite parts, (0 1? 2 ). 

Thus in the triangle ABC whose right angle is C ; 



M. 


Adjacent 


Parts. 


Opposite Parts. 


a, one of the sides including the 
right angle. 


f-*' 


b. 


7T . TT 

As c. 

2 2 


7T 

A, the complement of an angle. 


b, 


7T 
C. 

2 


2 


7T 

c, the complement of the hypothenuse. 


~-A 
2 A ' 


2 


a, b. 



We shall, whatever part be taken for M, prove Napier's rules, 
which are, 

sin M = Product of the tangents of the adjacent parts = tan A T . tanA 2 . 

sin M = Product of the cosines of the opposite parts = cos Oj . cos 2 . 



28 

I. Let a be the middle part. 



Then A l =--B, A 2 = b; O x ^--A, 2 = --e. 

1 2 2 2 



_, , cos B + cos A . cos C cos B 

NOW COS O = ; ; = — — , 

sin A . sin C sin A 

since in this case sin C = 1, and cos C = ; 



cos jB 

cos o = — 



sin a 
sin 5 . 



sin 6 
cos 5 sin b (ir 



• sin a 



sin 2? cos 6 



= tan (- -B\ .tan 6 (l). 



sin a sm A .. 

Again, — = - — — = sin J ; 

sin c sin 6 



'. sin a = sin A . sin c - cos I A ) . cos ( c J . . (2). 

ii. If b be the middle part, we have as in the last case, 
sin b = tan I A ) .tan a (l)« 

sin b = cos I 5 J .cos I c ) ....(2). 



7T 

in. Let c be the middle part. 



ThmA^- ~A, A t m--Bi { = a r 0,-6; 

2 2 



29 

cos C + cos A . cos B cos A cos B 

NOW, COS C = ; ; = - . — ; 

sin J . sin i? sin ^4 sm B 

•'■ sin (f ~ c ) = tan (f ~ A \ tan (f " s ) -' (1) ' 

cos c — cos a . cos 6 
Again, cos C = : ; — = i 

sm fl.smo 

.*. cos c = cos a . cos b ; 
.". sin I c) = cos a . cos 6 (2). 

7T 

iv. Let ^f be the middle part. 

Then A, = b, J 2 = - - c ; O l = --B, 2 =a; 

2 2 

cos c cos C + cos ^4 . cos B sin J. . sin C cos A 

^ow ■ = . = * 

cos b sin A . sin 5 cos i? + cos A . cos C sin 5 ' 

cos c . sin 6 cos c sin b cos c 

.*. cos A = sin 2? . = sm C. — — . =- = — - . -. — ; 

coso sine cos 6 cos 6 sine 

.*. sin ( A ) = tan b . tan I c) (l). 

A T> ^ A Sm I -^ ) 

cos A + cos 5 . cos C cos ^ \2 / 

Again, cos a = : — - — r— = - — 5 = ; 

sm B . sm C sm B (it \ 

• cos B) 

\2 / 

.-. sin (- - J J = cos I- - Bj . cosa (2). 

v. In like manner if - - B be the middle part, we have 

2 

sin ( B\ = tana, tan ( c) (l), 

sin ( - - B\ = cos (~ - Jj.cosfr (2), 



30 

39- To shew that in a triangle ABC, in which C is a 
right angle, A and a are of the same affection, as are also 
B and b. 

t-i . 7 * tan a 

Jb or sin 6 = cot A . tan a = . 

tan A 

Now since 6 is less than it, sin fr is positive ; — therefore tan a 
and tan A must be of the same sign. And because tt is the 
limit both of a and of A, these angles must be both greater or 
both less than a right angle ; that is, A and a are of the same 
affection. 

Similarly, from sin a — — , it appears that B and b are 

of the same affection. 

40. If in a right-angled triangle, an angle and the side 
opposite to it be the only given quantities, the triangle cannot 
be determined. Fig. 14. 

For if the circles AB and AC intersect again in A', and C 
be a right angle, it is evident that ACB and A'CB have the 
angles A, A' equal, and CB the side opposite to these angles 
is the same in both triangles. It is therefore ambiguous whether 
ABC or A'BC be the triangle sought. 

This ambiguity also appears in determining the triangle 
from Napier^ rules. For we cannot tell from the equation 

sin AC = tan CB . tan I A ) , 

whether we are to take the angle AC or its supplement A'C. 

41. The solutions from two given parts of the other cases 
of a right-angled triangle are not ambiguous, if attention be 
paid to the two principles, 

The greater side is opposite to the greater angle. 

An angle and the side opposite to it are of the same affection. 



31 

For example : Let c and A be given, to find «, B, b. 

Sin a = cos ( A ) . cos I - — c J = sin A . sin c. 

And since a and J are of the same affection, the greater or 
lesser angle which satisfies this equation is to be taken for a, 

according as A is greater or less than — . 



Again, sin I c I , or cos c, = cot A . cot B ; 

. * . tan B = cot A . sec c. 

And B is 90° as the second member of the equation is 

positive or negative; that is, as A and c are of the like or 
unlike affection. 

Again, sin I c J , or cos c, = cos a . cos b ; 

.*. cos b = cos c . sec«. 



< 7T 

And b is — , according as the second member of the 

equation is positive or negative, that is, as a and c are of the 
like or unlike affection. 

42. In selecting a formula, attention must be paid to the 
principles laid down in Appendix n. to PI. Trig. The following 
formulae may be used with advantage when the side or angle 
required is small, or nearly equal to one or to two right 
angles. 

By Napier's rules, cos c = cot A . cot B, whence c cannot be 
accurately determined, if it be either a very small angle or 
nearly equal to two right angles. 



Now 2 



32 

_ cos A . cos B 



r 


u— 1 f 

2/ 


— i — u 


uo t; 


— A — tUl 


-« . tut i» — J. — 


sin ^ . sin 


s' 






. c 
.*. sm- 

2 


Vv 


cos (A + B) 
sin ^1 . sin B 






So 


c 
cos — 

2 


= a/i 


K» 


+ cos c) 


/ cos (A 

^ 2 sin J . 

sin g - j) 
sin i? 


sin B 




In like manner 


from cos a = 


we have 





=v 



sin 
a 
cos 



in {+(B -A) + -\. cos k(S + 4) - ^ 



2 sin 5 



a y sinjj^ + ^-^.cos^-^} 



sin 

2 sin 5 



So from cos .4 = cot c . tan 6, 

.A .. ^ / sin (c - 6) 
sin — = a/4-(1 - cot c . tan 6) = \/ — — , 



2 sin c . cos b 



A . ^ / sin (c + b) 

cos — = a/ -i-(l -f cot c . tan 6) = \/ — 

2 v ' v 2smc.cos6 



When c and .4 are given, if a be nearly a right angle, it 
cannot be accurately determined from its sine. In this case 
cot B = cos c . tan A determines B, and a may be found from 

• a a 

the formula 1 lor sin — or cos - . 



33 

43. Def. A triangle is called quadrantal, if any one 
of its sides be a quadrant. 

To apply Napier s rules to the solution of quadrantal 
triangles. 

Let ABC be a quadrantal triangle, and let the side AB, 
or c, be a quadrant. Taking the polar triangle, we have 

for its angles, w — a ; it — b ; it — c, or — , 

for its sides, 7r — A; it — B ; ir — C. 

This triangle is therefore a right-angled triangle, the right 
angle being that opposite to the side 7r — C, and may be solved 
by Napier's rules ; its circular parts being, 



J 



(x - a), or a , 



for the angles ; < 












[;-<- 


-b), 


or b - 


7T 

2 ' 




(tt-A. 








for the sides, 


TT- B. 










2 v 


-c), 


or C- 


7T 
2 ' 



That is, the quadrantal triangle can be solved by Napier's 
rules, if we neglect the quadrantal side, and take these quan- 
tities for the circular parts. 



*E 



CHAPTER IV. 



ON THE SOLUTION OF OBLIQUE ANGLED TRIANGLES. 



44. Let the three sides be given, (a, b, c.) 

The angles may be determined from one of the formulae 
(xvi.), (xvii.), (xviii.), (xix.). 

45. Let the three angles be given, (A, B, C.) 

The sides may be determined from one of the formula? 
(xx.), (xxi.), (xxii.), (xxiii.). 

46. Let two sides and the included angle be given, 
(a, C, b). 

By Napier's first and second analogies, (xii.) and (xiii.), 

cos-±-(« - b) C 

tan \{A + B) = £ jr-™t-, 

cos \ (a + b) 2 

, / * ™ sm ir( a - b) C 

tan i (A - B) = -r-f-7 r( . cot - , 

2 v ' sin-i-(«+ b) 2 



{A + B) and \{A — B) are determined 

' *• are known. 



■ \B = 4-(^ + 



BU 
B)-i(A-B),j 



And A and B being known, c is found from 

sin C 



sm c = sin a . 



sin ^ 



35 



47. To determine c independently of A and B ? by forms 
adapted to logarithmic computation 

_. ^ cos c — cos a . cos b 
Cos C = : =— ; 

sin a . sin o 

.-. cos c = sin a . sin 6 . cos C + cos a . cos b 

= cos b . (sin « . tan b , cos C + cos «). 

Let $ be an angle such that 

tan 9 = tan b . cos C (l). 

_. _ , . sin 

Then cos c = cos b . (sin « . j- cos a) 

cos# 7 

cos 6 / 
= 5 .cos(o-0) (2). 

COS0 N 

From (l), L tan = L tan 6 + L cos C - 10 ; which gives 0. 

(2), Zcos c=Z, cos 6+ L cos (a — 0)— LcosO ; which gives c. 

48. Ze£ fo#o angles and the included side be given, 
(A, c, B). 

From Napier's third and fourth analogies, (xiv.) and (xv.), 

cos ±(A-B) c 



tan \{a + b) = 



cos ±(A + B)' 2 



7X sin ^(A-B) c 
tan i-(a -b) = - — -,— — .tan - , 

2V ; s\n \(A + B) 2 

t(« +6) and -§-(« — 6) are determined. 

■ • V? w z\ i / z.\ f are k "own. 
\b=jr(a + b) -±.(a-b),j 

And a and b being known, C is found from 



. sin c 

sin C = sin J . 

sin a 



36 



49. To find C independently of a and b, by forms 
adapted to logarithmic computation. 



Cos c 



cos C + cos A . cos B 



sin A . sin B 
,\ cos C = cos c.sinA.smB— cos ^4 . cos B ; 



.-. 2 cos 



— ) - 1 = 1 1 - 2 I sin - I > . sin A . sin 5 - cos A . cos 5 
= - cos (A + B) - 2 ( sin - ) . sin A . sin B ; 



,\ 2 ( cos — ) = 1 - cos ( A + B) - 2 ( sin - J . sin A . sin 5 ; 



•. cos 7 \ = { S m±(A+B)Y-- 



I . sin A . sin i?. 

9 



ess 



Now (sin - I . sin A . sin B is necessarily positive and 1 
than unity. Let therefore be an angle such that 

(sin 0)~ = I sin - j . sin A . sin B (l) ; 

.'. (cos|) 2 ={sini(^ + J B)} 2 -(sin^ 
= sin {±(A+B)+0}.sm{jr(A+B)-0}.FLTrig. Art. 47.. ..(2). 



From (l), LsinO = L sin- +i.Lsin J + -i-.Lsin 5-10 
which gives 0. 



FYom(2),Lcos- = i.Lsm{i(A+B) + Q\+i.Lsm{±(A+B)-0}i 
which gives C. 



37 

50. Let two angles and a side opposite to one of them be 
given, (A, B, a). 

c- * sin B 

bin b = sin a . — . 

smJ 

^, / .. n C cos ±-(a - b) J J ^ 

From (xn.), tan - = ~ '-. cot UA + B). 

v n 2 cos£(a+6) 2V 

From (x.v.), tan- = ^^ _ ^ ■ tan+(« + 6). 

And 6 having been determined from the first of these equations, 
C and c may be found from the two last. 

51. To determine C and c independently of b, by forms 
adapted to logarithmic computation. 

cos J + cos B . cos C 

Cos a = : — - — ;-— ; 

sin B . sin 6 

.*. cos A = cos a . sin 2? . sin C — cos 5 . cos C 

= cos B . (cos a . tan B . sin C — cos C). 

Let # be an angle such that 

cot = cos a . tan B (l); 

„ /cos0 . ^ \ 

.*. cos A - cos B . — — - . sm C - cos G 
Vsm / 

-sH-*<«-« «. 

From (l), L cot = L cos a + Z, tan B — 10 ; which gives 0. 

(2), Z sin (C - 6) = L cos ^ + Z, sin - Z, cos B ; which 
gives C — 0, and thence C. 



38 

Again, we have from (vii.), 

cot a . sin c — cot A . sin B + cos c . cos B ; 
.*. cot A . sin 5 = cot a . sin c — cos c . cos B 

= cot « . (sin c — cos B . tan a . cos c), 
whence, if be such that 

tan 9 = cos B . tan a, (l), 

we have 

. . _ cot a . . 

cot J . sin B - . sin (c - 6) (2). 

cos y 

From (l), L tan = L cos J5 + L tan « - 10 ; which gives 0. 

(2), L sin (c-0)=Z cot J+Z, sin B+L cos 0-£cot a-10 ; 
which gives c — $, and thence c. 

52. Let two sides and an angle opposite to one of them 
be given, (a, b, A). 



Sin B = sin A . 



sin b 



tan — — 



sin a 
C cos-|-(a — b) 



tan — = 



2 cos i(« + b) 
c cos±.(A + B) 



. cot 4- (-4 +5), 
. tan-i-(« + 6). 



2 cos-±-(J-#) 

And jB having been determined from the first equation, C and c 
are found from the two last equations. 

53. To determine C and c independently of B, % forms 
adapted to logarithmic computation. 

From (v.) we have, 

cot A . sin C = cot a . sin ft - cos C . cos ft ; 



39 

.'. cot a . sin b = cot A . sin C + cos C . cos b 

/cot ^ . \ 

= cos b . . sin C ■ + cos C ; 

Vcoso / 

Whence, if be such that 

. cot J 
cot0 = (1), 

cos b ' 

we have 

sin (C + 0) =cota . tan b . sin (2), 

and from (l) and (2) we can determine and C 

Again, cos a = cos A . sin b . sin c + cos 6 . cos c 

= cos b (cos ^f . tan b . sin c + cos c) ; 

Whence, if 6 be such that 

cot = cos J. . tan b (l), 

we have 

. . cos a . sin 

sm(c + 6)= (2), 

' cos b 

And from these two equations we can determine and c. 

54. In Article 52., the triangle has been apparently solved 

when two sides and an angle opposite to one of them are given. 

It is however ambiguous in some cases whether in determining 

. _ _ . . sin A sin a , 

A or B trom the equation = , we ought to take an 

H sin B sin 6 to 



angle which is less than — , or its supplement. We shall now 



7T 

2 

proceed to distinguish the ambiguous cases from those which 
are not so. 



40 

Let the given sides be a and b ; of which b is the greater. 
Produce the side c into a great circle, Art. 14. Cor., and let 
the pole be P. Figs. 15, 16, 17- 

Let dPCD be the great circle through P and C, it there- 
fore cuts at right angles the circle whose pole is P. 

Let CB = CB' = a, CA = CA f = b, 

Then DB = DB' ; and CBA = CBA', 

Let Dq = 90° = dq; .*. q is the pole of Dd, and .*. Cq=90°. 
Since a is less than o, CB falls nearer to CD than Cil does. 
Also, since CD < CB ; .\ C#D < CZ># < 90° ; ,\ CBd > 90°. 
So CJ/><90°, and CAd>90°. 

i. Let B, the angle opposite the greatest of the given 
sides, be given. 

1. Let B be < 90°. 

Then since a < b ; .-. A < B < 90°. 
Wherefore A can be determined, and the triangle be found. 

2. Let B be > 90°. 

First, let b be > 90°; then ^4 falls on 9^*7. Figs. 15, 16. 

Now if DZ? be > dA, Fig. 15., there are two triangles 
CBA and CBA' which answer the given conditions. 

If DB be < dA there is but owe solution, viz. BCA ; Fig. l6\ 
(because BAdA' being > 180°, BAAC is not a solution that 
can be admitted,) and in this case CAB is < 90°. 



41 
Now if DB < dA, cos DB > cos dJ ; 
cos CB cos CA 



cos CD cos Cd 
- cos ft 



cos « 



> 



, by Napier's rules ; 

; or cos a > cos ( 180° -ft), 



cos CD cos CZ> 

.-. a < 180° - ft, or a + ft < 180°. 

Next; let 6 be < 90°, then A falls on ^D^. And since 
B is > 90°, there is but one triangle {ACB) which satisfies 
the given conditions, (for A'dBC is inadmissible since Z CA r B 
would be > z CBA') ; and A is here < 90°. Also in this case, 
as in the last, a + ft is < 180°. 

We may therefore say that when B is > 90° the triangle 
may be determined, {A being < 90°), if a ■ + b be < 180°. 

ii. Let A, which is opposite to the less side, be the given 
angle. 

1. Let A be < 90°, .*. CAD is the given angle. Figs. 15, \6. 
First ; let a be < 90°. Then B falls on qDq. 

Then if DB <dA, Fig. 16., there are tfwo solutions, ^C2? 
and ACB'. 

If iX# > d A, Fig. 15., there is but owe solution, viz. JC^. 

And in this case B, or C#J, is > 90°. 

Now DB>dA, if cos DB < cos dJ, 

cos a cos 6 

< 



cos CD cos Cd ' 
cos a < cos (180° -ft), 

If a > 180° - ft, or a + ft > 180°. 
*F 



42 



Next; let a be > 90°, then B, B' fall on qdq. Fig. 17. 
In this case ACB is the only admissible solution ; and here we 
have B, or CBd,>90°. 

Here^ as well as in the former case, a + b is > 180°, — and 
we may therefore say that when A is < 90° the triangle may 
be found, (2? being > 90°), if a + b be > 180°. 



2. Let A be > 90°. Then since 6 > «, 
and the triangle may be determined. 



B > A > 90° ; 



55. On the whole then we collect that two sides, a and b, 
(of which b is the greater), being given and an angle opposite 
to one of them, the triangle may be found in the following 
cases. 



Description of the 
given angle. 


Limitation. 


Solution. 


B < 90° 

B > 90° 




A < 90° 
A < 90° 


a + b < 180° 


A < 90° 
A > 90° 


a + b > 180° 


B >90° 
5 >90° 





56. Similar ambiguities exist in the determination of the 
triangle apparently solved in Art. 50. We shall not however 
take each case separately, but make use of the results recapitu- 
lated in the last Article, to shew when the triangle can be deter- 
mined from two angles and a side opposite to one of them. 

Let A, B be the given angles, of which B is the greater, 
and a, b the sides respectively opposite ; therefore b > a. 

Take the polar triangle. Two of its sides are 180° - A, 
and 180° - B, of which 180°-^ is the greater; and the angles 
respectively opposite are 180°-« and 180° -b, of which 180°- a 
is the greater. 



43 



Now if we can point out from Art. 55. when one of the 
angles 180° - a or 180° - b of the polar triangle can be found 
from the two given sides 180° - A and 180° - B and an angle 
opposite to one of them, we determine when one of the sides 
a or b of the proposed triangle can be determined from two 
given angles A and B and a side opposite to one of them. 

57. Taking therefore the four cases of Art. 55., we have 



From 1 , 



Description of the 
given angle. 



180° - a < 90° 
180°-a>90° 



180°- 6<90° 
180°- b>90° 



Limitation. 



J(180°-J) + 
\ or A 



{ 

(180 -#)<180°,1 
+ B>180° J 



f(l80°- A) + (l80°-B) > 180°, 



\ 



or A + B<180° 



Solution. 



180°-6<90°, 
or b > 90°. 

180°-&<90°, 
or b > 90°. 



l-80°-o>90°, 
or a < 90°. 

180°-a>90°, 
or a< 90°. 



Whence we form the following Table, which shews when a 
triangle can be determined from two angles A and B, of which 
B is the greater, and a side opposite to one of them. 



4, 



Description of the 
given side. 
• 


Limitation. 


Solution. 


a > 90° 
a < 90° 




b > 90° 
b > 90° 


A + B > 180° 


b > 90° 
b < 90° 


A + B < 180° 


a < 90° 
a < 90° 





44 

58. By letting fall a perpendicular from an angle upon 
the opposite side, an oblique-angled triangle may be divided 
into two right-angled triangles, which may in most cases be 
solved by Napier's rules. The same ambiguities will however 
arise in this method of solution as have been shewn to exist 
when this construction is not used for the determination of the 
triangle. 

59- Prob. To Jind the radius of a small circle described 
about a given triangle in terms of the angles of the triangle. 
Fig. 18. 

Bisect AC and CB in D and E, and draw at right angles 
to AC and CB the arcs DP and EP intersecting in P. Join 
PA, PB, PC. 

Then cos PC = cos PD . cos DC = cos PD . cos DA = cos PA ; 

.-. PA = PC. Similarly, PB = PC 

Therefore P is the pole of the circumscribing circle. 

Let PA = p. Now cos PBE = cot PB . tan BE = cot p . tan - 



.-. cot p - cos PBE . cot - . 

r 2 

And since PAC, PBC, PAB are isosceles triangles, 

.-. 2/.PBE + 2/.PAD + <zaPAB = A + B+ C = 26" 

.-. z PBE = S' - (PAD + PAB) * $' - A. 

41 , , ... a /cos OS" - B). cos (S'-C) 

Also by (xxn.), cot - = V —. ---. — - 

J 2 v - cos .$" . cos (S' - A) 

/ cos (S' - A) . cos (S' - B) . cos (S' - C) 

... , t p = v r^ 



MHM 



45 

60. To determine the radius of the circumscribing circle 
in terms of the sides of the triangle. Fig. 18. 

Let PC = ,0; iPCA = 0; .-. lPCB=C-Q. 

b sin b 

Then cos = cot p . tan - = cot p . , 

r 2 r 1 + cos b 

a 1 — cos a 

cos (C - 0) = cot p . tan - = cot p . — : ; 

v r 2 ' sin a 

1 - cos a 1 + cos 6 cos (C - 0) _ . „ J n 

.-. . — : = - = cos C + sin C • tan tr 

sin a sin b cos 

cos c — cos a . cos b . . 

= j : h sin C . tan ; 

sin a . sin b 

.-. (sin a . sin 6 . sin C . tan 0) 2 
= { (l - cos a) . (l + cos 6) - (cos c - cos cr . cos 6) }% 

and (sin a . sin 6 . sin C) 2 = (sin a) 2 , (sin 6) 2 . { 1 - (cos C) 2 } 

= (sin a) 2 , (sin 6) 2 -(cos c— cos « . cos b) 2 ; 

Therefore by adding these two equations, we have, 
(sin a .smb. sin C) 2 .\ 1 + (tan 0) 2 } , or (sin a . sin b . sin C . sec 0) 2 , 
(sin a) 2 , (sin b) 2 + (l -cos a) 2 . (l + cos b) 2 — 2 (l -cos a) (l +cos b) (cos c— cos a . cos b) 
l-cosa).(l+cos&).{(l+cosa).(l — cos6) + (l— cos a). (1+ cos 6)— 2 (cose— cosa.cosb)l 
2(1 - cos a) . (l + cos b) . (l - cos c) 
6 



lo. sin- . cos - . sin - 

\ 2 2 2/ 



. « b . c 

4 . sin - . cos - . sin - 

2 2 2 

SeC 9 = ; ; ; i -— 

sin a sm b . sin C 



46 

b 
tan - 

2 h 

.'. tan p = - = tan - . sec 

r cos 6 2 

.a b , c 

4 . sin - . cos - . sin - 

b 2 2 2 

= tan - . = r— : — — - 

2 sina. sin b . sin C 

.a . b . c 

2 . sin - . sin - . sin - 

= ^/{sinS.sm(S-a).sm(S-b).sm(S-c)\ i y ' X1X )* 

61. To find the radius of a circle inscribed in a given 
triangle in terms of the sides of the triangle. Fig. 19. 

Bisect z A and Z C by AP and CP which meet in P, and 
from P draw PD, PE, PF perpendicular to the sides. 

Then sin PD = sin DCP . sin CP = sin PCE . sin PC = sin PE ; 

.-. PE = PD. Similarly, PF = PD. 

Therefore P is the pole of the inscribed circle. 

(j 

Let PD - p. Now sin CE = tan p . cot — ; 
r r 2 

.*. tan p = sin CE . tan — . 

r 2 

And since the angles A, B, C are bisected, it is clear that 
AD = AF, FB = BE, EC=CD; 

.-. 2 C£ + 2,1^ + 2^5 = a + b + c = 2#; 

.\ CjE = S - (AF + F#) = S - c. 

A1 , , ... C /sin (£ - o) . sin (S - b) 

Also, by (xvm.j, tan - = \/ : — . ; 

J v ' 2 v sin *y . sin (S - c) 

.'. tan p 



/sin (tf-a) . sin (S-b) . sin (^-c) 



sin S 



47 

62. If it be required to determine the radius of the in- 
scribed circle in terms of the angles of the triangle, let 

CD = 9, and .-. DA = b -0. 

From the equations 

C A 

sin = tan p . cot — , and sin (6 - 0) = tan p . cot — , 

we can prove in a manner similar to that employed in 
Art. 5Q., that 

ABC 

2 cos — . cos — . cos — 

P = ^/{-cosS f .cos(S f - A), cos (S' -B). cos (S'-C)}' 



CHAPTER V. 



ON THE AREAS OF SPHERICAL TRIANGLES, AND THE SOLUTION OF 

TRIANGLES WHOSE SIDES ARE SMALL COMPARED WITH 

THE RADIUS OF THE SPHERE. 



63. Def. The portion of the surface of a sphere which 
is intercepted between two great semicircles, as ACB and 
AEB, is called a lune. Fig. 20. 

64. To find the area of a lune. Fig. 20. 

If ACBD, ABBE, AEBF be lunes having each the 
same angle at J, it is clear that they are all equal, and if 
the angle CAD be repeated any number of times, the area 
ACBDA will be repeated the same number of times; 

i. e. the area of a lune varies as its angle ; 

area of lune whose angle is A A 

area of a sphere (whose angle is S60°) 360 ' 

And area of a sphere = iiirr 2 , if r — radius of the sphere ; 

(Hymers' Integral Calculus, p. 153.) 



•. area of lune whose angle is A = . 47rir 

* 360 



A 

180 



49 



65. To find the area of a spherical triangle. Fig. 21. 

Let ABC be a triangle upon a hemisphere ABBE, 
Art. 13. Cor., and let AC, BC, be produced till they meet 
again in F, which is on the side of the sphere turned from 
the spectator. 

Then since CDF = a semicircle = ACD, .-. DF = CA. 
Similarly, aDFE = +ACB; 
.'. A DFE = A ACB in every respect. 

Now 2= area of A ABC 

= surface of hemisphere - BHDC - AGEC - DCE 

= 2Trr 2 -(\uneAHD-?)-(\uneBGE-?)-(luneCDFE-I,) 

2-n-r* | 1 +32; by Art. 64. 

180 180 180/ J 



.-. 2 = (A + B + C-180 ). 
E 



wr 



180 



. 7rr 2 ; 



180° 
if £ = J + B + C-180 . 



Con. Hence for all triangles described on the same sphere, 

Soci + 5+C- 180°. 

And on this account A -f B + C — 180° has sometimes been 
taken as the measure of the surface of a triangle. 

Def. The quantity A + B + C - 180° by which the sum 
of the degrees in the angles of the spherical triangle exceeds 
180°, is called the spherical excess of the triangle. 

* G 



50 



66. Cagnoli's Theorem. To shew that if E be the 
spherical excess, 

E_ y/jsin S . sin (S - a) . sin (£ - b) . sin (ff - c)\ 



then sin 



a 6 c 
2 cos - . cos - . cos - 

2 2 2 



sin - = sin \{A + B+C- 180) = sin {±(A + B) - i(l80°- C) 



C C 

= sin-i-(4 + B) sin cos-U.4 + B) cos-. 

2 y 2 

c 

COS — 

2 

And by (viii.), sin ±(A + B) = . cos \(a - b) 

c 
cos - 

2 



sin — 
2 
(ix.), cos\(A + B) = . cos i(a + 6) 

cos- 

2 



c c 

n sin — .cos— 
.-. sin -= {cos^-(a - b) - cos \{a + 6)} ? 



cos 



2 



. a . b 

sin - . sin - 

2 2 



c 
cos - 

2 



• sinC (1) 



. a . b 
sin -. sin - 

2 2 2 



7—r- a/ \ sin ,S'. sin (#-a\ . sin (S-b) . sin (S - c) I 

sina.sin6 v L x ' v ' v " 



cos 



<y/ { sin S . sin (S - a) . sin (# - 6) . sin (S - c) \ 



a b c 
2 cos - .cos -. cos- 

2 2 2 



(2). 



51 



67- Llhuillier's Theorem. To shew that 

tan - = y/ {tan ±S . tan \{S - a) . tan \{S - 6) . tani(£ - c)} 

4 



By PI. Trig. Art. 45, 

sini(J+#)-sini(l80°-C) _ sin ±(A+B+C -180°) _ E 
cos^(J + jB) + cos-i-(l80°-C) ~ cosi(^ + #+C-180°) = an 7 ' 



(7 

£ sini(A + J B) -cos- 

.-. tan — = - 

4 . C 

cos \(A + B) + sin — 

At 



c C 

cos -|-(a - 6) - cos - cos — 

Z At 



c . C 
cos 4-(a -f b) + cos - sin — 



, by (viii.) and (ix.) 



smi(c+a-&).sini(c+&-a) / sin S. sin (S-c) 

cos±-(a+b+c).cos±(a+b-c) ' ^ sin (S-a). sin (5-6) 



= A /{tani*S'.tani(>S'-a).tani(^-6).tan-i-(^-c)}; 

by expressing the sines under the radical in terms of the sines 
and cosines of half the angles. 

68. There are some other expressions for — which we 

will deduce, though they are not, like Cagnolfs and Lhuillier's, 
adapted to logarithmic computation. 



52 



Cos — = cos {i(A + B) - +(180° - C)} 



c c 

= cos I (J+B) . sin - + sin \{A + B) . cos - . 



Jcos-i- («+&).( sin — j + cos-|-(a-6). (cos— ] >. By (ix.)and(viii.) 

cos— J 



. {cos-f(a + 6).(i -cosC) + cos -J- (a - b) . (l + cos C)} 



2 cos 



{[cosi(a-6)+cos-|-(a+6)] + cosC[cosi(a-6)-cosi(a+fe)]} 



2 cos 



1 f a b ^ • a . b ) 1 v 

= . <cos -.cos- + cos C. sin- . sin -> (1) 

c [ 2 2 2 2j v ' 

1 f a 6 cos c — cos a . cos 6 .a .61 

— . <cos- . cos- -\ ; : . sm - . sin -> 

c [ 2 2 sin a. sin 6 2 2j 

. <4 . I cos- J . ( cos- ) + cos c - cos a . cos b\ 



cos 



cos — 

2 



cab 
4 cos - . cos - . cos - 

2 2 2 



a b c 
4 cos - . cos- . cos- 

2 2 2 



{(l + cos a) (l + cos 6) + cos c - cos a . cos 6} 



1 + cos a + cos b + cos c 

a b c 

4 cos - . cos - . cos - 

2 2 2 



■(*)• 



Dividing (l) of this Article by (l) of Article 62. 



a b _, 

„ cot— . cot - + cos C 
E 2 2 

cot— = : — 

2 sm C 



(3). 



53 



Dividing (2) of this Article by (2) of Art. 66 9 
1 + cos a + cos b + cos c 



E 

COt = 

% ^{sm 5.sin(*S' 



a). sin (S - b). sin [S - c)} " ' Y* 

69. In the following Articles we shall suppose x 9 y 9 % to 
be the lengths of the arcs opposite to the angles A 9 B 9 C of the 
spherical triangle ABC described on a sphere whose radius is r, 
and whose sides are small compared with the radius of the 
sphere. Also, A' 9 B' 9 C are the angles of the plane triangle 
whose sides are w 9 y 9 %; and A v B X9 C x are the angles which 
are observed for the angles of ABC 9 the errors a, /3, y being 
committed at the respective observations ; so that 

A = A, + a, B = B 1 + fi 9 C = C x + 7. 

70. The area of a triangle whose sides are small com- 
pared with the radius of the sphere being approximately 
known, required the number of seconds in its spherical 
excess. 

Let the triangle be described on the surface of the Earth, 
which we will suppose to be a sphere. 

Let r = number of linear feet in the Earth's radius, 

number of square feet in the area of the triangle 

E 

" 180 

irr 



n 



Now 



irr 



7rr 2 . Art. 64 ; 

.-. n= (E. 60.60). 
length of an arc of 1° 



radius 



180. 60. 60 

1 
180 



7T. PI. Trig. Art. 90. 



— ■ =length of 1°, in feet, = 60859* 1 x 6, by actual measurement; 

loO 



180 

.60859-1 x6; 

7T 



n = (E . 60 . 60) . 



180 



7T . 60 . 60 



. (60859-1 x 6f 



= (£.60.60). 180 - (6085 - 91 > 2 
3-141592 



54 

.-. l 1Q (E.60. 60) = l w n- (l l0 180 + 2/ 10 6085'91 - l lQ 3-141592) 
= l 10 n - 9S267736 (l), 

where E .60. 60 is the number of seconds in the spherical 
excess. 

This reasoning being general, it is clear that the number 
of seconds in the spherical excess may always be determined 
from 



n = (E. 60.60). — — , 

v ' 180.60.60 



if the radius of the sphere be known, whatever be its mag- 
nitude. 

Cor. We may determine n in the following manner. 

n = area of spherical triangle in square feet. 

= area of plane triangle whose sides are a?, y, *?, and 
angles are the observed angles A 19 B 19 C x ; nearly. 

, „ sin B x . sin C x TO m . 
= iy.s.on J„ or ° & «* dn (Bi + ^ • P1 - T "g- Art *- 81 > 82 - 



General Roy in the Trigonometrical Survey of England 
approximately determined the area of the triangle to a suf- 
ficient degree of accuracy by laying down on paper the base 
and the observed angles at the base, and measuring the per- 
pendicular from the vertex on the base by the compasses on 
a scale. 

71. Ex. If the observed angles and a side x of the 
triangle ABC be 

Aj = 42°, 2', 32" j 

B 1 =67, 55, 39 > =179°, 59', 59". 

C, = 70, 1, 48 ! 

x = 27404-2 feet ; 

required the number of seconds in the sum of the errors made 
in observing the angles. 



55 



Here the apparent spherical excess is 

179°, 59', 59"- 180°= - l". 
Now a? being the side opposite to the angle A ] 
sin Z?] . sin C x 



2 ' " sin (B l + Cy 



and Z 10 .E.6o.6o = L 



{„ sin jBi . sin CA 
i'X'-^rh M\ ~ 9-32677.36 
2 sin (#, 4- Cj)J 



= U 



sin 67° 



„ ,'°, 55, 39 x sin 70°, l', 48 ] „ „ 

L (27404-2) 2 . ' ' . ' * } ^— [-9"3267736. 

v y sin 137°, 57 , 27' J 



Add, 2.^ 27404-2 =2x4*4378172= 8-8756344 

Z, sin 67°, 55', 39" = 9*9669434 

L sin 70°, l', 48" = 9*9730685 



28-8156463 



Subtract^ 2 = '3010300 

Lsinl37°,57 / 5 27 ,/ 
or L sin 42°, 2', 33" = 9*8258684 
10 
9-3267736 



29'4536720 

T-3619743 = / 10 ('23). 

And therefore the computed spherical excess, which may 
be supposed to be the real spherical excess, is •23". 

Hence it appears that the whole error of observation, viz. 
real spherical excess — apparent spherical excess, is -23" — (— l"), 
or l"-23, which has to be added to the three observed angles 
A l9 B l9 Cj, in such proportions as the observer's judgement 
directs. (See the next Article.) 



56 



72. To shew how the observed angles of a spherical tri- 
angle whose sides are small compared with the radius of the 
sphere may be best freed from the errors of observation. 

Let A, B, C be the real angles of the triangle, 

A l9 B l9 C l the observed angles, 

«? /3? 7 the errors made in observing A 9 B 9 C respectively. 

So that A + B + C = {A, + a) + (B l + fS) + (C, + 7 ) ; 

.-. (A+B + C- 180°) = (A, + B l +C l - 180°) + (« + j3 + 7) ; 

.-. a + /3 + 7 = (J + J B+C- 180°) - (J x + B l + C l - 180°) 

= real spherical excess - computed spherical excess. 

Now if, by the method of the last Article, we approximate 
closely to the real spherical excess, we can find by this equation 
the sum of the errors of observation which have been made. 
The distribution however of this sum, that is, the determining 
what part of the whole error is to be assigned to each angle 
individually, must evidently be left to the judgement of the 
observer, who from knowing the state of the atmosphere at 
the times of observation, may guess how far the observations 
may be depended on, and thence may to each of the observed 
angles assign the whole error in what proportion he thinks 
proper. 

If by the method of Art. 70, a + /3 + 7, the sum of the 
errors of observation, be found, and thence the angles A, B, C 
be determined by the arbitrary assignment of the several parts 
of this whole error to each of the observed angles A x , B Y , C l9 
the sum of the errors of observation may be supposed to be got 
rid of. Yet it is highly probable that the judgement of the 
observer has not been absolutely correct in this arbitrary assign- 
ment of the parts of the whole error. The following theorem 
will point out what relation the sides of the triangle ought to 
bear to each other, in order that the small quantities by which 
the corrected angles differ from the real angles of the triangle, 
may have the least possible effect in producing errors in the 
determination of the other two sides of the triangle from a 
measured side and the corrected angles. 



57 

73. Having given the corrected angles and one side of 
a spherical triangle whose sides are small compared with the 
radius of the sphere, required the relation which the sides 
of the triangle ought to bear to each other, in order that 
the other sides may be determined from these data with the 
least possible amount of error. 

Let x, y, % be the sides opposite to the real angles of the 
triangle A, B, C ; a, /3', y, the errors of the corrected angles ; 
therefore A + a, B + ft, C + y are the corrected angles. 

Now since the spherical excess, and therefore the sum 
of the real angles of the triangle, is supposed to be known 
exactly, the sum of the corrected angles is known exactly, 
and therefore the sum of their errors a, /3', y must vanish ; 

.-. a' + ^ + y^O; .«. a' = - (/3' + 7'). 

(We shall throughout this Article neglect powers of a, /3', y' 
of any order above the first.) 

Now, considering the spherical triangle as very nearly a 
plane triangle, 

sin (C + 7') 



% - x . 



= x 



sin (A + a) 

sin (C + y) 

sin {J -03'+ 7')} 

sin C . cos y + cos C . sin y 
sin A . cos (/3' -1- 7') - cos A . sin (/3' + 7') 

sin C 1 + y' . cot C . 



sm 

sin C 
sin A 

sin C 



1 +7 , .cotC}.{l + 03' + 7 r ).cot^ 



= ,v 



. {l + y . (cot A + cot C) + 6'. cot 4} 

sm A *■ 

sin C [ , sin C sin (J + C) a , cos J. sin CI 

sin A \ ' sin A sinyl.sinC (sin J) 2 J 

*H 



58 

Now sin B = sin (?r - i?) = sin {A + C), nearly. 

Therefore the error in the value of %, or z — w— — -, is 

sin A 

{. sin B n . cos A . sin C] , . 

^.•(Suy^- (sin^r } (1) - 

Similarly the error of t/ is 

. iff Si " C | -/ COsJ - si " g l . (2) 

The question now reduces itself to this, viz. to determine what 
values of A, B, C make the quantities (l) and (2) the least 
possible. 

Now since a + /3' + y = 0, two of these quantities are of 
the same sign and one of a different sign. Wherefore the 
probability is that /3' and y are of different signs. If /3' and y 
be of different signs, the expressions (l) and (2) are diminished 
in magnitude by giving cos A a positive value*, that is, 
by supposing A to be less than a right angle. 

If we make this further supposition, that the errors 
/3' and y' though different in sign are yet nearly equal in 
magnitude, it is clear that (l) and (2) satisfy this hypothesis, 
if B be nearly equal to C. 

The conclusion therefore we arrive at is, The greatest 
probability of correctness in solving a small spherical triangle 
from three observed angles and a measured side, is when the 
angle opposite to the known side is less than a right angle, and 
the other two sides are nearly equal. And these conditions 
will be best fulfilled, for a series of triangles, if care be taken 
that each triangle be nearly equilateral. 

* It is evidently more advantageous in determining a series of triangles from one 
another, that the errors should be equally diffused through all, than that any one 
calculated side should by differing much from its real value, affect with great errors 
all the triangles successively determined from it. If, /3' and y' being of different signs, 
cos A become negative, the errors (1) and (2) are increased, and considerable errors 
may be thus introduced into the calculations. 



59 

This Article has been amplified from Art. 179- of the 
" Trigonometry ," by Professor Airy, published in the Ency- 
clopedia Metropolitana, to which treatise the student is referred 
for a short and clear explanation of the principles of Geodetic 
measurements, 

74. It may be as well to recount the gratuitous sup- 
positions which have been made in the last Article. 

I. The spherical excess has been accurately determined. 

II. The errors /3' and y are of different signs. 

in. These errors are of the same magnitude. 

The reasoning therefore in the last Article is evidently 
not very exact. For a particular triangle it is very probable 
that some, or all of these conditions may not be fulfilled. But 
since our reasoning is on general principles, we may be certain 
that the results obtained in the determination of a great 
number of triangles will be more nearly correct when the 
triangles are all nearly equilateral, than when they are of any 
other form. 

75. We shall not here enter into the details of geodetic 
operations : we will however demonstrate the following methods 
of determining a spherical triangle whose sides are small com- 
pared with the radius of the sphere. 

76. Legendre's Theorem. If each of the angles of 
a spherical triangle whose sides are small when compared with 
the radius of the sphere be diminished by one third of the 
spherical excess, the triangle may be solved as a plane triangle 
whose sides are equal to the sides of the spherical triangle, 
and whose angles are these reduced angles. 

Let x 9 y, % be the lengths of the sides respectively opposite 
to the angles A, B, C, of a small spherical triangle described 
on the sphere whose radius is r. Let A\ B\ C' be the angles 
of the plane triangle whose sides are a?, y, %. 



60 



ItjA _|_ a*« rp~ 

Now cos A = — , 

2yz 

and cos A . sin b . sin c = cos a — cos 6 . cos c, 

. 2/ • % x V % 

or cos A . sin - . sin - = cos cos - . cos - . 

ix* /y ry> if y 

Expanding the sines and cosines of these small angles 
by Art. 101, PI. Trig., and neglecting powers of the angles 
above the fourth, as we shall do throughout this investiga- 
tion,- — (there will therefore be no powers or products of <#, y, z 
of more than four dimensions), — we have, 



cos A 



x* \ I f y* \ f z 2 z 4 \ 

.3Ar 4 ) " V "^ + 2.3.4rV ' V ~ 2r 2 + 2.3Ar 4 ) 



(i _ yi \ (- _ ** ) 

\r 2.3r d ) ' \r 2.3r 6 ) 

1 v 2 z 2 1 
— (y 2 + z 2 - at 2 ) -- 1 —, + 4 . (^-s/ 4 -* 4 ) 

2r 2 J 2.2r 4 2.3 Ar 4 ' 

V*_ / f + * 8 Y 

r 2 '{ 2.3r 2 ) 

1 [2 y 4 + 4>y 2 z 2 + 2 z 4 - 2 x 2 y' - 2 a? VI 

Ayzr r \ + # 4 -y 4 - z 4 - 6y 2 z' 2 J 



2y 



y 2 + z 2 — x 2 



+ 



2yz 2.3, 

= cos A' -. \2oc 2 y 2 -\-2w 2 z 2 + 2y 2 z 2 — oc 4 — y l — z*\ 

2.3Ayzr 2 l y y * 

(2zyf < '»•* ' ~ 8 - ^ 2 ^ 



2.3Ayzr 



ty~ + z 2 - x 2 \ 2 \ 
{ 2~y~z ) J 



= C os A - -— . { l - (cos AY} ; 
2.3r l 



yz 



7 . (sin Ay=cob A - cos A=2s\n4, (A-A).*m -i (^ + ^') 



61 

[Now the angles A and A are in practice expressed in 
degrees ; when expressed by the circular measure they become 

A A' 

. 7T and . 7T, PI. Trig. Art. 90. Also A = A nearly. 

180 180 & J 

A- A' A -A' 

.-. sin-UJ - A ) = sin 4-. . tt = 4-. . 7r- 

2V / 2 180 180 

PI. Trig. Art. 102. 

And sin \{A + A') = sin A f nearly. 

Also yx . sin J' = 2AA'B f C = 2AABC (nearly) = 2. — . Trr 3 . 

180 

Making these substitutions our equation becomes,] 

A ^ A' 

. 7r . sin A ; 



2 


. — . 71-r 

180 


■sin J' = 


2.1 


J -A 




2 . 3r l 


180 






E 

3 ' 


= J - 


-A; 






.'. ii' 


= J 


E 

3 



So B' = B-- , and C = C - - . 

3 3 

If then BC or w be measured on the sphere and the angles 

A, B, C be observed, by subtracting one third of the spherical 

excess from each of these angles, and solving the plane tri- 

E E E 
angle AB'C' from the angles A , B , C and 

3 3 3 

the given side <#, we get AB and AC, — that is, y and x, the 
other arcs forming the sides of the spherical triangle ABC. 

Cor. If powers of the angles above the second be not 
admitted, we have, 

. V ~ 2W V irV * \ " 2r 2 J y 2 +x 2 -w 2 

cos A — ■ — = = cos A . 

y x %yx 

r r 



62 



77- To find the angle contained between the chords of 
two spherical arcs which subtend given angles at the center 
of the sphere, the angle between the arcs themselves being also 
given. Fig. 22. 

Let AB and JC be the arcs, the center of the sphere. 
Let the lines AO, AB, AC meet the surface of a sphere which 
is described with center A and any radius AD in the points 
D, E, F respectively. Then the angle EDF is the inclination 
of the planes BAO and CAO, i. e. the angle contained between 
the arcs AB and AC, or zi. 

T AB AC , 

Let — — = c, — — = o. 
AO AO 

Now aCA0 = \(tt~ A0C) = \{ir-b). 
So lBA0 = \(tt-c). 
And from the triangle EDF, 

cos EF = cos DF . cos DE + sin DF . sin DE . cos EDF ; 



Or cos EAF= sin 



• c be 
sin - + cos - . cos - . cos A . (1). 



78. We will now deduce from the result obtained in the 
last Article, two formulae which are convenient for determining 
practically the angle EAF. 

Let EAF = A-0; 

Then cos (A -0) -cos A = sin- .sin - + (cos- .cos J J .cos A; 

v J 2 2 V 2 2 / 

And, PL Trig. Art. 47, 

. b . c ( . b + c\ 2 ( . b - c\ 2 

sin -.sin- = sin — sin , 

2 2 \ 4/V 4 ) 



cos- .cos- 

O 9 



/ b+cY I . b-c\ 2 I . b+c\ 2 I . &-c\s 

l cos — ) - l s,n T-j ='-r— ) - r~r) 



63 
Therefore our equation becomes 



2 sin [ A ~ !) • sin H ( sin b -ir ) V ( sin hr) 1 
"{( sin ^ c ) + ( sin6 ~r- c )}' cosJ; 



. 6 

'. 2 sin - 
2 



1 f / . b + c\ 2 , ^ ( . b~c\ 2 , t 1 

0\ '\V ln T"j •( 1 - cos ^)-( v sm '-^-j .(H-cos^)L.(i). 



sin \ A — I 



a 
For an approximation to the value of 0, make sin - = - , 



and sin ( A — ) = sin A ; 



2 (sin— j 
rp , . 1 - cos A V 2 / J 

I hen, since — — — - = . = tan — , 

sin A .A A 2 

2 . sin — . cos — 
2 2 

_ 1 + cos A A 

and — : — = cot — ; 

sin A 2 

_ / . 6 + c\ 2 A ( . b-c\ 2 A 

.'. 6 - \ sin . tan sin . cot — ; 

\ 4 / 2 V * 4 / 2 

Therefore the number of seconds in 6 

A 1 / . b-c\ 2 A 

. ( sin . cot — ; 

2 sml" V 4 / 2 



l / . b + c\ 

sm . tan 



sin 1 sin 1 



And by determining the two terms of the second member of 
this equation separately by means of tables of logarithms, we 
obtain the number of seconds in 0, and thence may determine 
the angle A - 0, which is contained by the straight lines BA 
and CA. 



64 

79- The next formula we shall deduce from the equation 
of Art. 77? involves the lengths of the arcs AB and AC. 

Let r be the radius of the sphere in feet, and y, % be the 
lengths of the arcs AC and AB in feet. 

b . c b c 
Now cos (A — 0) - cos A = sin - . sin - + (cos - . cos 1 ) . cos A ; 

v ' 2 2 2 2 \ 

And cos (A-6) -cos A =2 sin f^f 1 .sin- = 2 sin A. sin -nearly, 

. b . * # _ 

sin - = sin — = — nearly, 
2 2r 2r J 



b % -. sr 

- = cos — = l - i — 

2 2r "* 4r^ 



cos - = cos -— = l - 1 — - nearly 



.-. 2sinJ.sin- = — .-£- + J (l -A.-^r) ■ [i-4-tt) -1>.cosJ; 
.-. . sin J = - — — . {4<yz -2(y 2 + %*) . cos A} 

"^•{ &+ * ),, v(™t)" (sf " jr)1 - 2 ( COi i)} ; 

. ■ /y +af\2 J /« -#\ 2 A 

• • 0= .tan- - (" ) .cot-; 

V 4r / 2 V 4r / 2 . 



A A 

tan — cot 



, n v y + %V 2 (y-%\ 2 2 

.-. seconds in = - — - = Z . _ — - £ . ; 

sin l" V 4r / sin l" V 4r / sin l" 

And the value of these terms being separately calculated by 
logarithms, the number of seconds in may be determined; 
and A having been found by the method of Art. 71, EAF 
may be found. 



65 



If then the radius of the Earth and the sides of the spheri- 
cal triangle be approximately known, the chordal triangle can 
be determined, since its sides can be found by the expression, 

chord = 2 r . sin \ — - , and its angles can be determined by 
A rad. 

one of the two last Articles. A series of such chordal triangles 

is available, by methods which we shall not here explain, to 

determine the figure and situation of a country on the Earth. 

80. A third method of performing geodetic measurements, 
is to solve the triangles by the rules laid down for the solution 
of spherical triangles whose sides are not small in comparison 
with the radius of the sphere. In this case the logarithms of 
the sines and tangents of the sides, which are very small, must 
be found by the methods pointed out in Appendix m to 
PL Trig. 

81. Each of these three methods has had its followers. 
Very great accuracy was attained in the English survey by 
using the chordal triangles. The method of Legendre was 
employed in measuring an arc of the meridian in France. 
The third method is that preferred by Delambre. 



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